^  LIBRARY 

- 

UNIVERSITY  Ot  CALIFORNIA. 

'  Deceived  WAR  15JR93       .  i8g    . 

Accessions-  No. 


forks  of  Professor  Mansfield  Merriman. 


Published   by  JOHN  WILEY  &  SONS,  53  E.  Tenth 
Street,  New  York. 


A  TREATISE  ON  HYDRAULICS. 

Designed  as  a  Text- Book  for  Technical  Schools  and  for  the 
use  of  Engineers.  By  Professor  Mansfield  Merriman,  Lehigli 
University.  Fourth  edition,  revised 8vo,  cloth,  $3  50 

"As  a  whole  this  book  is  the  most  valuable  addition  to  the  literature  of 
hydraulic  science  which  has  yet  appeared  in  America,  and  we  do  not  know 
of  any  of  equal  value  anywhere  else." — Railroad  Gazette. 

"With  a  tolerably  complete  knowledge  of  what  has  been  written  on 
Hydraulics  in  England,  France,  Germany,  United  States,  and  to  some  extent 
Italy,  I  have  no  hesitation  in  saying  that  I  hold  this  book  to  be  th^  best 
treatise  for  students,  young  or  old,  yet  written.  It  better  presents  the 
primary  essentials  of  the  art."— From  CLEMENS  HERSCHEL,  Hydraulic 
Engineer  of  the  Holyoke  Water  Power  Company. 

A  TEXT-BOOK  ON  THE  METHOD  OP  LEAST  SQUARES. 

By  Mansfield  Merriman,  C.E.,  Ph.D.,  Professor  of  Civil 
Engineering  in  Lehigh  University.  Fifth  revised  edition. 

8vo,  cloth,  2  00 

This  work  treats  of  the  law  of  probability  of  error,  the  ad- 
justment and  discussion  of  observations  arising  in  surveying, 
geodesy,  astronomy  and  physics,  and  the  methods  of  compar- 
ing their  degrees  of  precision.  Its  rules  and  tables  will  assist 
all  who  wish  to  make  accurate  measurements. 

"This  is  a  very  useful  and  much  needed  text-book." — Science. 

"Even  the  casual  reader  cannot  fail  to  be  struck  with  the  value  which 
such  a  book  must  possess  to  the  working  engineer.  It  abounds  in  illustra- 
tions and  problems  drawn  directly  from  surveying,  geodesy  and  eugineer- 
ing."—  Engineering  Ntws. 

THE   MECHANICS  OF  MATERIALS  AND   OF  BEAMS, 
COLUMNS,    AND    SHAFTS. 

By  Professor  Mansfield  Merriman,  Lehigh  University,  South 

Bethlehem,  Pa. 

Fourth  edition  revised  and  enlarged.     8vo,  cloth,  interleaved,  3  50 

"We  cannot  commend  the  book  too  highly  to  the  consideration  of  all 
Professors  of  Applied  Mechanics  and  Engineering  arid  Technical  Schools 
and  Colleges,  and  we  think  a  general  introduction  of  the  work  will  mark  an 
advance  in  the  rational  of  technical  instruction."— American  Engineer. 

"The  mathematical  deductions  of  the  laws  of  strength  and  stiffness  of 
beams,  supported,  fixed,  and  continuous,  under  compression,  tension  and 
torsion,  and  of  columns,  are  elegant  and  complete.  As  in  previous  books 
by  the  same  author,  plenty  of  practical  original  and  modern  examples  are 
introduced  as  problems."— Proceedings  Engineers'  Club  (/Philadelphia. 


A  TEXT-BOOK  ON  ROOPS  AND  BRIDGES. 

Being  the  course  of  instruction  given  by  the  author  to  the 
students  of  civil  engineering  in  Lehigh  University. 
To  be  completed  in  four  parts. 

PART!.     STRESSES    IN    SIMPLE    TRUSSES.     By  Professor 
Mansfield   Merriman.     Third  edition.     8vo,  cloth $250 

"  The  author  gives  the  most  modern  practice  in  determining  the  stresses 
due  to  moving  loads,  taking  actual  typical  locomotive  wheel  loads,  and 
reproduces  the  Phoenix  Bridge  Co's  diagram  lor  tabulating  wheel  move- 
ments. The  whole  treatment  is  concise  and  very  clear  and  elegant."— Rail- 
road Gazette. 

PART  II.     GRAPHIC  STATICS.     By  Professors  Mansfield  Mer- 
riman and  Henry  S.  Jacoby.      Second  edition.      8vo,   cloth,  2  50 

"  The  plan  of  this  book  is*  simple  and  easily  understood ;  and  a?  the  treat- 
ment of  all  problems  is  graphical,  mathematics  can  scarcely  be  said  to  enter 
into  its  composition.  Judging  from  our  own  correspondence,  it  is  a  work 
for  which  there  is  a  decided  demand  outside  of  technical  schools." 

— Engineering  News. 

PART  III.     BRIDGE  DESIGN.     In  Preparation. 

This  volume  is  intended  to  include  the  design  of  plate  girders, 
lattice  trusses,  and  pin-connected  bridges,  together  with  the 
proportioning  of  details,  the  whole  being  in  accordance  with 
the  best  modern  practice  and  especially  adapted  to  the  needs 
of  students. 

THE     FIGURE     OF     THE     EARTH.      An    Introduction    to 

Geodesy. 

By  Mansfield  Merriman,  Ph.D.,  formerly  Acting  Assistant 
United  States  Coast  and  Geodetic  Survey.  12mo,  cloth 1  50 

"  It  is  so  far  popularized,  that  there  are  few  persons  of  ordinary  intelli- 
gence who  may  not  read  it  with  profit  and  certainly  great  interest.  "—Engi- 
neering News. 

"A  clear  and  concise  introduction  to  the  science  of  geodesy.  The  book 
is  interesting  and  deals  with  the  subject  in  a  useful,  and  to  some  extent, 
popular  manner." — London  Engineering. 

A  TEXT-BOOK  ON  RETAINING  WALLS  AND  MASONRY 
DAMS. 

By  Professor  Mansfield  Merriman,  Lehigh  University. 

8vo,  cloth,  2  00 

This  work  is  designed  not  only  as  a  text -book  for  students 
but  also  for  the  use  of  civil  engineers.  It  clearly  sets  forth 
the  methods  of  computing  the  thrust  of  earth  against  walls, 
and  the  investigation  and  design  of  walls  and  dams  in  the 
most  economic  manner.  The  principles  and  formulas  are 
illustrated  by  numerous  numerical  examples. 


A   TEXT-BOOK 


ON 


RETAINING  WALLS 


AND 


MASONRY  DAMS. 


BY 

MANSFIELD    MERRIMAN, 
«j 

PROFESSOR  OF  CIVIL  ENGINEERING  IN  LEHIGH  UNIVERSITY. 


In  scientiis  ediscendis  prosunt  exempla  magis  quam  praecepta.  —  NEWTON. 


NEW   YORK: 
JOHN    WILEY    &    SONS, 

53,  EAST  TENTH  STREET. 
1892. 


COPYRIGHT,  1892, 

BY 
MANSFIELD  MERRIMAN. 


ROBERT  DRUMMOND,  FERRIS  BROS., 

Electrotyper,  Printers, 

444  and  446  Pearl  St.,  326  Pearl  Street, 
New  York.  New  York. 


CONTENTS. 


CHAPTER  I. 
EARTHWORK  SLOPES. 

PAGE 

Article  I.  Equilibrium  of  Loose  Earth I 

2.  The  Cohesion  of  Earth 4 

3.  Equilibrium  of  Cohesive  Earth q 

4.  Stability  of  Slopes  in  Cohesive  Earth 12 

5.  Curved  Slopes  and  Terraces 16 

6.  Practical  Considerations 21 

CHAPTER  II. 

THE  LATERAL  PRESSURE  OF  EARTH. 

Article  7.  Fundamental  Principles  24 

8.  Normal  Pressure  against  Walls 27 

9.  Inclined  Pressure  against  Walls 31 

**  10.  General  Formula  for  Lateral  Pressure 35 

**n.  Computation  of  Pressures. .....  36 

*•«  12.  The  Centre  of  Pressure 39 

CHAPTER  III. 
INVESTIGATION  OF  RETAINING  WALLS. 

Artie!  e  13.  Weight  and  Friction  of  Stone 42 

14.  General  Conditions  regarding  Sliding. ...    44 

15.  Graphical  Discussion  of  Sliding 47 

16.  Analytical  Discussion  of  Sliding 50 

17.  General  Conditions  regarding  Rotation 54 

18.  Graphical  Discussion  of  Rotation 57 

19.  Analytical  Discussion  of  Rotation 59 

20.  Compressi ve  Stresses  in  the  Masonry 63 

iii 


IV  CONTENTS. 

CHAPTER  IV. 
DESIGN  OF  RETAINING  WALLS. 

*       PAGE 

Article  21.  Data  and  General  Considerations 68 

22.  Computation  of  Thickness. ...   71 

23.  Security  against  Sliding , 75 

24.  Economic  Proportions 77 

25.  The  Line  of  Resistance. 83 

26.  Design  of  a  Polygonal  Section 86 

27.  Design  and  Construction 90 


CHAPTER  V. 
MASONRY  DAMS. 

Article  28.  The  Pressure  of  Water 93 

29.  Principles  and  Methods 95 

30.  Investigation  of  a  Trapezoidal  Dam 98 

31.  Design  of  a  Trapezoidal  Section 103 

32.  Design  of  a  High  Trapezoidal  Section 107 

33.  Economic  Sections  for  High  Dams ...  109 

34.  Investigation  of  a  Polygonal  Section 112 

35.  Design  of  a  High  Economic  Section 115 

36.  Additional  Data  and  Methods 120 


SLOPES,   WALLS   AND    DAMS 


CHAPTER   I. 
EARTHWORK    SLOPES. 

ARTICLE  i.    EQUILIBRIUM  OF  LOOSE  EARTH. 

Earthwork  slopes  are  the  surfaces  formed  when  excava- 
tions, embankments,  terraces,  mounds,  and  other  construc- 
tions are  made  in  or  with  the  natural  earth.  The  earth  is  to 
be  regarded  in  discussion  as  homogeneous  and  inelastic,  and 
as  consisting  of  particles  more  or  less  united  by  cohesion  be- 
tween which  friction  is  generated  whenever  exterior  forces 
tend  to  effect  a  separation.  As  some  kinds  of  earth  when 
dry  are  destitute  of  cohesion,  these  will  first  be  considered 
under  the  term  "  loose  earth." 

The  friction  of  earth  upon  earth  will  be  taken  to  be  gov- 
erned by  the  same  approximate  laws  as  for  other  materials, 
namely:  first,  the  force  of  friction  between  two  surfaces  is 
directly  proportional  to  the  normal  pressure ;  second,  it  varies 


2  EARTHWORK  SLOPES.  [CHAP.  I. 

with  the  nature  of  the  material ;  and  third,  it  is  independent  of 
the  area  of  contact.  These  laws  may  be  expressed  by  the 
equation 

F=fN, (i) 

in  which  N  is  the  normal  pressure,  F  the  force  of  friction 
perpendicular  to  N,  and  f  is  a  quantity  called  the  coefficient 
of  friction  which  varies  with  the  kind  of  material.  As  F  and 
N  are  both  in  pounds,  f  is  an  abstract  number  ;  its  value  for 
earth  ranges  from  about  0.5  to  i.o. 

If  a  mass  of  earth  be  thoroughly  loosened  so  as  to  destroy 
all  cohesion  between  its  particles,  and  then  be  poured  verti- 
cally upon  the  point  D  in  the  horizontal  plane  BC,  it  will  form 
a  cone  BAG,  all  of  whose  elements  AB,  AC,  etc.,  make  equal 


o 


FIG.  i. 

angles  with  the  horizontal.  This  angle  ABC  is  called  the 
"  angle  of  repose,"  or  sometimes  "  the  angle  of  natural  slope," 
and  it  is  found  by  experiment  that  eacli  l.ind  of  earth  has  its 
own  constant  angle.  The  particlepof  earth  on  such  a  slope 
are  held  in  equilibrium  by  the  forces  of  gravity  and  friction. 
Let  <p  be  the  angle  of  repose -A13D,  and  f  the  coefficient  of 
friction.  In  the  figure  draw  W  vertically  to  represent  the 
weight  of  a  particle,  and  let  N  and  F  be  its  components  nor- 


• 

ART.  i.]  EQUILIBRIUM  OF  LOOSE  EARTH.  3 

mal  and  parallel  to  the  slope.     Now  since  motion  is  about  to 
begin, 

F  =  fN. 

Also  since  the  angle  between  A7"  and  J^is  equal  to  the  angle 
of  repose  0,  the  right-angled  triangle  NO  Ogives 


Therefore  results  the  important  relation 

/=  tan  0,    ........     (2) 

that  is,  the  coefficient  of  friction  of  earth  is  equal  to  the 
tangent  of  the  angle  of  repose.  It  is  hence  easy  to  determine 
/when  0  has  been  found  by  experiment. 

In  building  an  embankment  of  joose  earth  it  is  necessary 
that  its  slope,  or  angle  of  inclination  to  the  horizontal,  should 
not  be  greater  than  the  angle  of  repose.  When  making  an 
excavation  it  is  often  possible,  on  account  of  the  cohesion  of 
the  earth,  to  have  its  slope  at  first  greater  than  the  angle  of 
repose,  but  as  the  cohesion  disappears  under  atmospheric  in- 
fluences the  particles  roll  down  and  its  slope  finally  becomes 
equal  to  0. 

The  following  table  gives  rough  average  values  of  the 
angles  of  repose  and  coefficients  of  friction  of  different  kinds 
of  earth.  In  the  fourth  column  the  inclination  or  slope  is  ex- 
pressed in  the  manner  customary  among  engineers  by  the 


EARTHWORK  SLOPES. 


[CHAP.  I. 


ratio  of  its  horizontal  to  its  vertical  projection.     In  the  last 
column  average  values  of  the  weight  of  the  material  .are  given. 


Kind  of  Earth. 

Angle 
of 
Repose. 

* 

Coefficient 
of 
Friction. 

f 

Inclination, 
cot  0 

Weight. 

Kilos  per 
cu.  met. 

Pounds 
per  cu.  ft. 

Gravel    round  

30° 
40 

35 
40 
30 
40 
45 
32 

0.58 
0.84 
0.70 
0.84 
0.58 
0.84 
1.  00 
0.62 

.7  to 
.2  to 
.4  to 
.2  to 
-7tO 
.2  tO 

to 
.6  to 

1600 
1700 
1600 
1700 
2000 
1440 
1520 
1840 

100 
IIO 
100 
IIO 

125 
90 

95 
H5 

Gravel    sharp.  ...       .....     . 

Sand,  dry  

Earth    dry  

Earth    moist           .    .               . 

It  will  be  noticed  that  the  natural  slope  and  specific 
gravity  of  earth  undergo  quite  wide  variations  as  its  degree 
of  moisture  varies.  In  collecting  data  for  the  discussion  of 
particular  cases  it  is  hence  necessary  to  determine  limits  as 
well  as  average  values. 

Problem  i.  A  bank  of  loose  earth  is  16  feet  high,  and  its 
width,  measured  on  the  slope,  is  28  feet.  Compute  the  co- 
efficient of  friction  and  the  angle  of  repose. 


ARTICLE  2.    THE  COHESION  OF  EARTH. 

Cohesion  is  a  force  uniting  particles  of  matter  together. 
If,  for  instance,  two  surfaces  have  been  for  some  time  in  con- 
tact, they  become  to  a  certain  extent  glued  or  fastened  to- 


ART.  2.]  THE   COHESION  OF  EARTH.  $ 

gether  so  that  any  attempt  to  separate  them  is  met  by  a 
resistance.  Friction  only  resists  the  separation  of  surfaces 
when  motion  is  attempted  which  produces  sliding,  but  cohe- 
sion resists  their  separation  whether  the  motion  be  attempted 
parallel  or  perpendicular  to  the  plane  of  contact.  Particles  of 
rock  are  held  together  by  strong  cohesive  forces,  while  parti- 
cles of  dry  sand  have  little,  if  any,  cohesion. 

By  experiment  the  following  are  found  to  be  the  laws  of 
cohesion :  first,  the  force  of  cohesion  between  two  surfaces  is 
directly  proportional  to  the  area  of  contact ;  second,  it  depends 
upon  the  nature  of  the  surfaces ;  and  third,  it  is  independent 
of  the  normal  pressure.  These  laws  may  be  expressed  by  the 
equation 

C=cA (3) 

in  which  C  is  the  resisting  force  of  cohesion  between  two 
surfaces,  A  the  area  of  contact,  and  c  a  quantity  called  the 
coefficient  of  cohesion  depending  upon  the  nature  of  the 
material. 

The  value  of  c  for  homogeneous  earth  may  be  found  as 
follows  :  Dig  in  the  ground  several  trenches  of  considerable 
length  compared  with  their  width,  and  of  different  depths. 
After  a  few  days  it  will  be  observed  that  all  those  over  a  cer- 
tain depth  have  caved  along  some  plane  such  as  BM  in  Figure 
2.  Let  H  be  the  value  of  this  certain  depth.  Let  w  be  the 
weight  of  a  cubic  unit  of  earth,  and  0  its  angle  of  repose  when 


EARTHWORK  SLOPES.  [CHAP.  I. 

devoid  of  cohesion.     Then  the  coefficient  of  cohesion  of  the 
earth  may  be  computed  from  the  expression 

_  Hw(  i  —  sin  0) 
4  cos0 

This  formula  will  now  be  demonstrated. 

Let  the  plane  BM  in  the  figure  make  an  angle  x  with  the 
horizontal.'  The  prism  BAM,  whose  length  perpendicular  to 
the  drawing  will  be  taken  as  unity,  tends  to  slide  down  the 

M 


# 


FIG.  2. 


plane.  Let  W  be  the  weight  of  this  prism,  and  P  and  N  its 
components  parallel  and  normal  to  BM.  P  tends  to  cause 
motion  down  the  plane,  and  this  is  resisted  by  the  combined 
forces  of  friction  and  cohesion,  acting  in  the  plane.  The  force 
of  friction  is  fN,  and  that  of  cohesion  is  c/}  if  /  be  the  length, 
or  area,  of  BM.  At  the  moment  of  rupture 


from  which  the  value  of  c  is 


c  = 


P-fN 


(4) 


v 


, 

Rput*    ,  «  ^  '  JW***^    ^r^ 

A     :  ,]  T/inSSSm^  WE$m™+ 

y      «Sd^U^Sxt^    Jl^X^^  r       ZCOU^ 

In  order  to  determine  the  angle  ^r,  consider  that  the  equations 
just  written  are  for  the   plane  of   rupture,  and  that  for  any 


P  — 
other  plane./5  is  less  than  fN-\-  cl,  or  -  -  ---  is  less  than  c. 

'  l 

The  value  of  x  for  the  plane  of  rupture  is  hence  that  which 
renders 

~~T~~  ~  a  maximum  ......    (5) 


Equation  (5)  will  determine  the  value  of  x,  and  then  c  will  be 
found  from  (4). 

To  do  this  insert  first  for  P  its  value  W  sin  x,  for  N  its 
value  £Fcos  x,  and  for  fits  value  tan0.     Then  (5)  becomes 

£^&  ffi**j[X  |N»  '  r*'A*"')L 

'*'  w 

——(smx  —  tan  0  cos^)  =  a  maximum. 


- 
This  may  be  written 


W  sin  (x  —  0) 

-  -,  --  -  —  •  =  a  maximum. 
/cos  0 

Next  express  W  in  terms  of  H  and  ^r,  and  the  weight  of  a 
unit  of  volume  of  earth  w.     Thus 


and  then  (5)  becomes 

sin  (90°  —  x)  sin  (x  —  0) 

2  COS0 


=  a  maximum. 


7 


Z'-  *2^. 

.'  X  /^r2    V    f^?  / 

8  EARTHWORK  SLOPES.  [CHAP.  I. 

This  expression  is  a  maximum  when  the  two  variable  factors 
'      *:    are  equal;  or  when 


*-*  =  *-•  ^ 

Thus  the  value  of  x  for  the  plane  of  rupture  is 


-  jY    .    (6) 

Now  to  find  <:,  insert  in  formula  (4)  the  values  of  P,  N,  and 
f  in  terms  of  x,  and  it  becomes 

Hw  sin  (90°  —  x)  sin  (#  —  0) 

'  =  ~  2COS0  ~~> 

•which  by  virtue  of  (6)  reduces  to 


£  =    

Since  sin8  (45  —  -J-0)  equals  J(i  —  sin0),  this  value  becomes 
Hw(i  —  sin  0) 

C  =^  ' ,    ........       (7) 

4  cos  0 
which  is  the  formula  that  was  to  be  demonstrated. 

From  this  formula  the  numerical  value  of  c  can  be  com- 
puted when  H,  w,  and  0  are  known.  For  earth  weighing  100 
pounds  per  cubic  foot  and  whose  angle  of  natural  slope  is  30 
degrees,  the  value  of  c  becomes  i^.^H.  If  the  vertical  ruptur- 
ing depth  H  is  one  foot,  c  is  14.4  pounds  per  square  foot ;  but 
if  H  is  ten  feet,  then  c  is  144  pounds  per  square  foot. 


ART.  3.] 


EQUILIBRIUM   OF  COHESIVE   EARTH. 


Problem  2.  A  certain  bank  of  earth,  which  has  a  natural 
slope  when  loose  of  1.25  to  i,  stands  by  virtue  of  its  cohesion 
with  a  vertical  face  when  H  =  3  feet.  If  this  bank  fails,  find 
the  slope  immediately  after  rupture. 


ARTICLE  3.    EQUILIBRIUM  OF  COHESIVE  EARTH. 

If  the  particles  of  earth  be  united  by  cohesion,  a  slope  may 
exist  steeper  than  the  angle  of  repose.  Let  Figure  3  rep- 
resent the  practical  case  of  an  excavation  ABC  whose  slope 
AB  makes  an  angle  6  with  the  horizontal  greater  than  the 


angle  of  repose  0.  AM  is  the  natural  surface  of  the  ground 
making  with  the  horizontal  an  angle  tf  less  than  0.  It  is  re- 
quired to  determine  the  relation  between  the  slope  6  and  the 
vertical  depth  h  in  order  that  rupture  may  just  occur. 

Let  BM  be  the  plane  along  which  rupture  occurs,  and  x  its 
inclination  to  the  horizontal.  The  weight  of  the  prism  BAM 
tends  to  urge  it  down  the  plane,  and  this  is  resisted  by  the 
forces  of  friction  and  cohesion  actinga»  thfi^lane.  Let  Wbz 

UNIVERSITY 


10  EARTHWORK  SLOPES.  [CHAP.  I. 

the  weight  of  the  prism  for  a  length  unity,  and  P  and  N  its. 
components  parallel  and  normal  to  the  plane.  P  is  the  force 
causing  the  downward  sliding,  fN  is  the  resisting  force  of 
friction,  and  cl  that  of  cohesion,  if  /  be  the  area,  or  length,  of 
BM.  At  the  moment  of  rupture  P  —  fN-\-cl,  which  may  be 
written  *P<tf+&£,/  fk 


Now  as  x  varies  the  forces,  P  and  N  vary;  and  for  any  other 

P  —  fN 
plane  except  that  of  rupture  --  ~-  -  is  less  than  c.     Hence 

the  condition  which  will  determine  the  value  of  x  is 


/fi          tw  ***  y  /£  ,        -^     inw* 

P-fN 


When  x  has  been  found  from  (9)  its  value  is  to  be  inserted  in 
(8)  and  thus  the  relation  between  0  and  h  be  established. 

To  do  this  insert  in  (9)  for  P  and  N  their  values  W  sin  x 
and  W  CQSX,  and  for /its  value  tan  ,0.     Then  jit  takes  the  form 


"U~<,  W  sin  (x  —  0) 

7 =  a  maximu 

/COS0 

The  value  of  W  is  the  volume  of  the  prism  BAM,  multiplied: 
by  the  weight  of  a  unit  of  volume  w,  or 


W  =  \BA  .  BM.  w  sin  (0  -  x)  =  \hlw  ^L_£i 

v 

P  «  •.    - 


ART.  3.]  EQUILIBRIUM  OF  COHESIVE  EARTH. 

and  hence  the  expression  becomes 
hw  sin  (0  —  x)  sin  (x  — 


II 


=  a  maxmum. 


2  cos  0  sin  0 
This  is  a  maximum  with  respect  to  x  when 

/i  j> 


that  is,  the  plane  of   rupture  bisects  the  angle  between  the 
lines  of  natural  slope  and  excavated  slope,  or 


Now  if  (8)  be  expressed  in  terms  of  xy  it  becomes 

hw  sin  (6  —  x)  sin  (x  —  0)  =  2c  cos  0  sin  0, 
and  by  virtue  of  (10)  this  reduces  to 

hw  sin8  J(0  —  0)  =  2£  cos  0  sin  0; 

and  substituting  for  sin2  £(0  —  0)  its  value  f(i  —  cos(0  —  0)), 
and  f£m  £  its  value  from  (7),  there  is  found 

£(  I  —  cos  (0  -  0))  =  H(\  -  sin  0)  sin  0,     .     .     (11) 

,  4Jfc  ^^0>J^\<s^^ 

and  this  is  the  equation  of  condition  between  h  and  0. 

This  discussion  shows  that  both  the  angle  of  rupture  x  and 
the  relation  between  h  and  0  are  independent  of  the  slope  d 
made  by  the  natural  surface  of  the  ground  with  the  horizontal. 


12  EARTHWORK  SLOPES.  [CHAP.  I. 

By  the  help  of  formula  (n)  the  limiting  height  h  may  be 
found  when  0,  0  and  H  are  given.  For  instance,  let  it  be 
required  to  build  a  slope  of  I  to  i,  or  #  =  45°,  and  let  the 
earth  be  such  that  0  =  30°  and  ff=6  feet.  Then  the  depth 
at  which  rupture  will  occur  is 


For  stability  the  depth  must  of  course  be  less  than  62  feet, 
and  precautions  be  taken  that  the  cohesion  of  the  earth  be  not 
destroyed  by  the  action  of  the  weather. 

Problem  3.  Let  a  bank  whose  height  is  30  feet  and  slope 
45  degrees  be  of  earth  for  which  0  =  34°  and  H  =  3  feet. 
How  much  higher  can  it  be  raised,  keeping  the  same  slope, 
before  failure  will  occur  ?  ^  5' 


ARTICLE  4.  STABILITY  OF  SLOPES  IN  COHESIVE  EARTH. 

0 
In  practice  it  is  desired  to  determine  the  slope  of  a  bank 

so  that  it  may  be  stable  and  permanent.  To  deduce  an  equa- 
tion for  this  case  consider  again  Figure  3,  and  let  BM  be  any 
plane  through  the  foot  of  the  slope  making  an  angle  x  with 
the  horizontal.  As  x  varies  the  forces  P  and  N  vary,  and  it  is 
easy  to  see  that  the  weakest  plane  is  that  for  which  the  expres- 

P —  fN  3 

sion  — -r- —  is  a  maximum.     As  in  Art/J,  the  value  of  x  ren- 


ART.  4.]     STABILITY  OF  SLOPES  IN  COHESIVE  EARTH.         13 

dering  this  a  maximum  is  x  =  j-(#  -|-  0).     Now  it  is  required 
that  rupture  shall  not  occur  along  this  plane,  hence 

and     P-fN<cl\ 


or  if  n  be  a  number  greater  than  unity,  called  the  factor  of 
security, 

n(P-fN)  =  cl.     ......     (12) 

Rupture  can  now  occur  only  when  the  weight  W  becomes  n 
times  that  of  the  prism  of  earth  above  the  weakest  plane.  To 
adapt  this  equation  to  practical  use  it  is  only  necessary  to  sub- 
stitute for  P  and  N  their  values  in  terms  of  x,  and  then  to  make 
x  equal  to  \(9  -f-  0).  The  substitution  is  performed  exactly 
as  before,  and  leads  to  the  following  result  : 

n  h(\  —  cos  (8  —  0))  =  H(i  —  sin  0)  sin  0,    .    .     (13) 


which  is  the  required  equation  of  stability. 

>£e^^^^  +&(4 

If  n  is  unity,  this  of  course  reduces  to  the  case  of  rupture 
as  given  by  (i  i).  The  value  to  be  assigned  to  the  factor  n  can 
only  be  determined  by  observation  and  experiment  on  existing 
slopes.  Probably  about  2  or  3  will  prove  to  be  sufficient. 

When  6  is  given,  the  value  of  h  is  derived  at  once  directly 
from  (13),  thus: 

,  _  H(i  —  sin  0)  sin  6 

~?z(i-cos(0-0))  ...... 


14  EARTHWORK  SLOPES.  [CHAP.  I 

This  shows  that  the  height  for  security  should  be  -  of  that 

n 

for  rupture.  Thus  it  was  found  in  Article  3,  if  H  —  6  feet,  and 
0  =  30°,  jthat  the  limiting  height  for  a  slope  of  45°  would  be 
62  feet.  Hence  with  a  factor  of  security  of  2  the  height  would 
be  31  feet,  and  with  a  factor  of  3  the  height  would  be  20  feet. 

When  h  is  given  and  0  is  required,  the  formula  for  stability 
may  be  written  in  the  form 

I  —  cos(<?  -  0)  _  H(i  —  sin  0) 
sin  6  nh 

The  second  member  is  here  a  known  quantity  and  may  be 
called  a.  By  developing  the  numerator  in  the  first  member 
and  then  substituting  for  sin  6  and  cos  6  their  values  in  terms 
of  tan£#,  a  quadratic  expression  results  whose  solution  gives 


This  determines  the  slope  6  for  a  factor  of  security  n. 

For  example,  let  it  be  required  to  find  the  slope  0  for  a 
bank  25  feet  high  with  a  factor  of  security  of  1.5,  the  value  of 
0  being  30°  and  that  of  H  being  t  feet.  Here 


«=       =  °-°667' 


ART.  4-]     STABILITY  OF  SLOPES  IN  COHESIVE  EARTH. 

and  then  from  the  formula, 


tan  \Q  —  0.304  -f-  V  —  0.0718  +  0.0922  =  0.447. 

Hence  %6  is  about  24°  and  0  is  about  48°,  or  a  slope  of  0.9  to  I. 
The  slope  when  built  must  of  course  be  protected  from  the 
action  of  the  weather  in  order  to  preserve  the  cohesion  of  the 
earth. 

The  security  of  a  bank  may  be  investigated  by  measuring 
its  height  h  and  slope  6,  and  finding  by  experiment  the  angle 
of  repose  <p  and  vertical  rupturing  depth  H.  Then  from  (13) 
there  is  found 

H(i  -sin0)sin0 
= 


For  example,  let  it  be  required  to  find  the  factor  n  when 
h  =  30  feet,  0  =  45°,  0  =  34°,  and  H=  3  feet.  Substituting, 

3  X  0.441  X  0.707 
30  X  0.0184 

If  such  a  slope  had  existed  many  years,  and  if  the  values  of  0 
and  //"were  the  most  unfavorable  that  could  occur,  it  might 
be  concluded  that  the  factor  of  security  deduced  is  sufficiently 
high  ;  but  if  such  a  slope  should  be  observed  to  fail,  it  would 
be  necessary  to  conclude  that  the  factor  is  too  low. 

Problem  4.  A  certain  slope  has  h  =  25  feet,  0  =  30°,  and 
H  =  5  feet.  At  what  angle  6  will  rupture  occur  ?  What  is  its 
factor  of  security  if  0  be  48  degrees  ? 


i6 


EARTHWORK  SLOPES. 


[CHAP.  L 


ARTICLE  5.    CURVED  SLOPES-  AND,  TERRACES. 

The  preceding  articles  clearly  show  that  the  angle  of  slope 
6  of  a  bank  of  cohesive  earth  increases  as  its  vertical  height  k 
decreases,  and,  conversely,  that  as  h  becomes  greater  0  be- 
comes smaller.  It  would  hence  appear  that  the  upper  part  of 
a  bank  may  be  steeper  than  the  lower  part,  and  its  liability  to 
rupture  be  the  same  throughout.  To  determine  the  form  of 


M 


a       d 


v-v'v'  .v-u< 


FIG.  4. 

such  a  curve,  let  D  in  the  figure  be  any  point  upon  it  whose 
ordinate  Dd  is  y.  Let  DM  be  the  weakest  plane  making  an 
angle  x  with  the  horizontal.  The  prism  of  which  aDM  is  a 
section,  by  virtue  of  its  weight  W,  tends  to  slide  down  the 
plane.  Let  P  and  N  be  the  components  of  ^parallel  and 
normal  to  the  plane.  If  n  be  the  factor  of  security,  the  condi- 
tion, as  in  Art.  4,  is 

n(P-fN)-cl=o. 

By  inserting  for  P,  N  and  /  their  values  in  terms  of  x,  this 
becomes 

nW(\  —/cot  x)  —  cy(\  +  cot*  x)  =  o. 


ART.  5.]  CURVED    SLOPES  AND    TERRACE s".  IJ 

The  value  of  W  for  a  prism  one  unit  in  length  is  found  from 
the  difference  of  the  areas  dDM  and  dDa,  or  if  A  represent 
the  surface  dDa, 

W  =  w^y*  cot  *  —  A). 

By  inserting  this  the  equation  of  stability  becomes 

cot*  —  A)  (i  —  /cot*)  —  cy(i -{- cot* *fy<=  °-     (17) 


This  expression  equals  zero  for  the  weakes>^lane,  but  for  any 
other  plane  its  value  is  less  than  zero.  Hence  it  must  be  a 
maximum  with  respect  to  x  or  cot  *,  and  its  first  derivative 
must  vanish.  Thus,  also, 


cot  x  —  A)  (  —  /)  +  nw(i  —f  cot  ^)i(/2)  —  zcy  co^  x  =  o.    (  1  8) 


By  eliminating  cot  x  from  (17)  and  (18),  the  following  value 
of  A  is  found  : 


A  =  ~nfWy  +  *c~2  V2^(»/a7  +  2^(i  +/"),    (19) 


and  this  is  the  practical  equation  of  the  required  curve,  A 
being  the  area  between  the  curve  and  any  ordinate  whose 
value  is  y. 

For  example,  let  it 'be  required  to  construct  a  curve  of 
equal  stability  in  a  bank  of  40  feet  height  with  a  factor  of 
security  of  1.5,  the  earth  having  a  natural  slope  of  31°,  a  verti- 
cal rupturing  depth  of  5  feet,  and  weighing  100  pounds  per 
Cubic  foot.  Here,  from  (2)  and  (7),  there  is  first  found 

f  =  0.6,     c  =  71  pounds  per  square  foot, 


1  8  EARTHWORK  SLOPES.  [CHAP.  I. 

and  formula  (19)  becomes 


A  = 


284-2  ^193(907  +142)]. 


From  this  are  computed  the  following  special  values  : 

For  y  =  10  feet  A  =    27  square  feet  ; 

For  y  =  20  feet,  A  =  159  square  feet  ; 

For  y  =  30  feet,  A  =  421  square  feet  ; 

For  y  =  40  feet,  A  =  809  square  feet. 

These  are  the  areas  between  the  slope  and  the  given  ordinate, 
and  may  be  practically  regarded  as  consisting  of  trapezoids, 
as  shown  in  Figure  5.  The  first  area  is  that  of  the  triangle 
abB,  hence 

£  .  10  .  ab  =  27,     or    ab  —  5.4  feet. 

The  second  area  comprises  the  triangle  AbB  and  the  trapezoid 
bBCc,  hence 


In  a  similar  manner  cd  •=•  10.5  feet  and  de  =  n.i  feet,  and  the 
four  points  B,  C,  D,  and  E  are  thus  located.  In  Figure  5  the 
portions  of  the  slope  are  drawn  as  straight  lines ;  it  may  be  so 
built,  or  intermediate  points  of  the  curve  be  established  by  the 
eye. 


ART.  5.] 


CURVED    SLOPES  AND    TERRACES. 


It  is  not  difficult  to  deduce  from  (19)  the  co-ordinate  equa- 
tion giving  the  relation  between  the  abscissa  and  ordinate  for 
every  point  of  the  curve,  but  it  is  of  such  a  nature  as  to  be  of 
little  practical  use.  In  the  manner  just  explained,  as  many 
points  upon  the  curve  may  be  located  as  required.  It  is  seen 
from  (19)  that  A  is  negative  for  small  values  of  y,  or  theoreti- 


A     al>  c     d      e 


FIG.  5. 

cally  the  curve  overhangs  the  slope.  Practically,  of  course, 
the  equation  should  not  be  used  for  values  of  y  less  than  //, 
and  it  will  usually  be  found  advisable  and  necessary  that  the 
upper  part  of  the  curve  should  be  reversed  in  direction  so  as 
to  form  an  ogee,  as  shown  by  the  broken  line  AB  in  Figure  5. 

When  terraces  are  to  be  constructed,  it  is  evident  that  the 
upper  one  may  have  the  greatest  slope  and  the  lower  one  the 
least  slope.  Formula  (19)  may  be  used  for  this  purpose,  since 
the  area  A  is  not  necessarily  bounded  by  a  curved  line,  but 
may  be  disposed  in  any  form  desired. 

For  example,  take  a  bank  30  feet  high  in  which  it  is 
desired  to  build  three  terraces,  as  in  Figure  6,  with  a  factor  of 


20 


EARTHWORK  SLOPES. 


[CHAP.  I. 


safety  of  1.5.     The  height  of  each  terrace  is  10  feet,  and  there 
are  two  steps  BC  and  DE,  each  4  feet  wide.     Let  w  =  100 


FIG.  6. 


pounds,  0=  31°,  and  //=  5  feet,  as  found  by  experiments. 
Then/"=  0.6  and  c  =  71,  and  formula  (19)  becomes 


A  = 


From  this,  when  y  =  10,  ^  =  27;  when  jj/  =  20,  ^4  =  159;  and 
when  y  =  30,  -4  =  421.  The  abscissas  are  now  found  to  be 
ab  =  5.4,  cd  —  6.1,  and  */"=  8.9  feet.  The  three  slopes  are 
hence  as  follows  : 


For  aB, 
For  CD, 
For  EF, 


SB  23         and 
10 


=•  -  and 

10 


=  ?2        and 


ART.  6.]  PRACTICAL    CONSIDERATIONS.  21 

To   insure  the   permanency  of   these   slopes   they  should   be 
protected  from  the  weather  by  sodding. 

Problem  5.  Design  a  terrace  of  four  planes,  the  upper  one 
being  6  feet  in  vertical  height,  the  lowest  10  feet,  and  the 
others  8  feet ;  the  steps  to  be  5  feet  in  width.  The  earth  is 
such  that  cot  0  =  1.5  to  i,  and  H  =  3  feet. 


ARTICLE  6.    PRACTICAL  CONSIDERATIONS. 

The  preceding  theory  and  formulas  can  be  usually  applied 
to  the  construction  of  embankments  as  well  as  to  excavations, 
provided  that  care  be  taken  to  compact  the  earth  to  a  proper 
degree  of  cohesion  and  the  slopes  be  protected  from  the  action 
of  the  elements.  The  height  h  is  always  given,  and  it  is  re- 
quired to  find  the  slope  0.  Unless  h  be  very  large  the  weak- 
est plane  will  intersect  the  roadway ;  but  if  not,  the  application 
of  the  formulas  can  only  err  on  the  side  of  safety.  The  load 
upon  the  roadway  can  be  regarded  as  a  mass  of  earth  uniformly 
distributed  over  it  and  thus  increasing  the  height  h. 

For  instance,  if  w  =  100  pounds  per  cubic  foot,  0  =  34°  and 
H=4  feet,  let  it  be  required  to  find  the  slope  0  for  an  em- 
bankment 30  feet  high.  For  security  the  weight  of  the  loco- 
motive should  be  taken  high,  say  6000  pounds  per  linear  foot 
of  track,  or  about  5CXD  pounds  per  square  foot  of  surface  for  a 
12-foot  roadbed,  which  would  be  equivalent  in  weight  to  a  mass 
of  earth  about  4  feet  "high.  Then  the  value  of  h  to  be  used 


22  EARTHWORK  SLOPES.  [CHAP.  I. 

in  formula  (15)  is  34  feet.     If  the  factor  of  security  be  2,  the 
value  of  a  is  0.0259,  and 


tan  £#  =  0.320  +  V  —  0.093 5+0. 1024  =  0.414. 

Hence  %0  is  about  22^  degrees  and  0  is  about  45°,  or  the  slope 
is  I  to  i.  The  proposed  embankment  with  this  slope  contains 
47  cubic  yards  per  linear  foot,  while  with  the  natural  slope  of 
34°  it  would  contain  62  cubic  yards  per  linear  foot.  A  saving 
in  cost  of  construction  will  hence  result  if  the  expense  of  pro- 
tecting the  slopes  to  preserve  the  cohesion  be  not  too  great. 

The  degree  of  moisture  of  earth  exercises  so  great  an  influ- 
ence upon  its  specific  gravity  and  angle  of  repose  that  special 
pains  should  be  taken  to  ascertain  the  values  of  those  quanti- 
ties which  are  the  most  unfavorable  to  stability.  In  general 
a  high  degree  of  moisture  increases  w  and  decreases  0.  These 
causes  alone  would  tend  to  increase  the  cohesion,  but  at  such 
times  H  usually  becomes  so  small  that  c  is  greatly  diminished. 
The  determination  of  H  is  awkward  and  there  seem  to  be  few 
recorded  experiments  concerning  it.  Care  should  be  taken 
that  the  trench  is  long,  or  that  transverse  cuts  be  made  at  its 
ends  so  that  lateral  cohesion  may  not  prevent  rupture,  and  a 
considerable  time  should  be  allowed  to  elapse  so  that  the  co- 
hesion may  be  subject  to  unfavorable  weather. 

• 

The  general  conclusions  of  the  above  theory  are  valuable, 

but  it  should  be  applied  with  caution  to  particular  cases,  not 
only  on  account  of  the  variability  in  the  data  but  on  account 
of  our  ignorance  of  the  proper  factor  of  security.  Numerical 


ART.  6.]  PRACTICAL    CONSIDERATIONS.  2$ 

computations,  however,  may  often  prove  useful  as  guides  in 
assisting  the  judgment.  As  shown  above,  a  great  saving  in  the 
cost  of  moving  material  will  result  if  slopes  be  built  in  accord- 
ance with  the  theory,  but  evidently  the  cost  of  properly  pro- 
tecting the  slopes  will  be  increased.  Should  the  latter  cost 
prove  to  be  the  smaller,  the  theory  will  ultimately  become  of 
real  practical  value. 

The  preceding  theory  is  not  new,  having  long  since  been 
set  forth  in  many  French  and  German  books,  but  the  author 
is  unaware  to  what  extent  it  is  practically  used  in  those*  coun- 
tries. The  introduction  of  a  factor  of  security  is,  however, 
believed  to  be  novel  in  this  connection,  and  by  proper  experi- 
ments for  determining  its  value  the  practical  application  of  the 
formulas  here  given  may  perhaps  be  rendered  possible. 

Problem  6.  A  railroad  cut  is  to  be  made  in  material  for 
which  w  =  100  pounds  per  cubic  foot,  @)=  32°,  ff  —  5  feet. 
If  h  is  40  feet  and  the  roadbed  16  feet  wide,  find  the  quantity 
of  material  necessary  to  excavate  when  the  slopes  have  a  fac- 
tor of  security  of  3.  ^4-.° 


24  THE  LATERAL  PRESSURE   OF  EARTH.       [CHAP.  II. 


CHAPTER  II. 

THE:  LATERAL  PRESSURE  OF  EARTH. 

ARTICLE  7.    FUNDAMENTAL  PRINCIPLES. 

A  retaining  wall  is  a  structure,  usually  nearly  vertical, 
which  sustains  the  lateral  pressure  of  earth.  In  investigating 
the  amount  of  this  pressure  it  is  generally  regarded  best  to  neg- 
lect the  cohesion  of  the  earth,  and  to  consider  it  as  loose 
(Article  i).  This  is  done,  partly  because  the  effect  of  cohesion 
is  difficult  to  estimate  and  partly  because  the  results  thus  ob- 
tained are  on  the  side  of  safety  for  the  wall, — the  entire  inves- 
tigation in  fact  being  undertaken  for  the  purpose  of  using  the 
results  in  designing  walls.  The  values  given  in  Article  I  for 
the  weight  of  earth  and  for  the  angles  of  repose  will  be  used 
in  this  chapter,  but  it  is  again  mentioned  that  they  are  subject 
to  much  variation,  and  that  in  practical  problems  the  values 
most  dangerous  to  stability  should  be  selected. 

The  pressures  against  a  retaining  wall  are  least  near  the 
top  and  greatest  near  the  base.  The  resultant  of  all  these 
pressures  is  called  the  "  resultant  pressure,"  or  simply  the  pres- 
sure, and  is  designated  by  the  letter  P.  The  determination  of 


ART.  7.] 


FUNDAMENTAL  PRINCIPLES. 


formulas  for  the  values  of  P  for  different  cases  is  the  object  of 
this  chapter. 

Let  the  resultant  pressure  P  against  the  back  of  a  wall  be 
resolved  into  a  component  N acting  normal, and  a  component 
.F  acting  parallel, to  the  back  of  a  wall.  Let  z  be  the  angle 
between  N  and  the  direction  of  P',  then 

F  =  N  tan  z. 

Let /be  the  coefficient  of  friction  between  the  earth  and  the 
wall,  then  for  the  case  of  incipient  motion, 

F=Nf. 

Therefore,  since  /is  the  tangent  of  the  angle  of  friction,  the 
angle  z  cannot  be  larger  than  the  angle  of  friction  between  the 


earth  and  the  wall  unless  the  earth  is  moving  along  the  wall. 
Various  views  are  held  by  authors  regarding  the  direction  of 
the  pressure  P,  or  the  value  of  the  angle  z.  Some  take  z  as 
zero,  or  regard  the  thrust  as  normal  to  the  wall ;  others  take  z 


26  THE  LATERAL   PRESSURE   OF  EARTH.         [CHAP.  II. 

as  equal  to  the  angle   of  repose  of  the  earth,  0 ;  while  a  few 
take  z  as  intermediate  between  these  values. 


In  Article  8  the  friction  of  the  earth  against  the  wall  will  be 
neglected,  or  the  angle  z  will  be  taken  as  o°.  The  value  of  the 
pressure  determined  under  this  supposition  will  be  called  the 
"normal  pressure/'  and  will  be  designated  by  Pr  It  is  not  to 
be  forgotten  that  the  actual  pressure  against  the  back  of  a  re- 
taining wall  cannot,  like  the  pressure  of  water,  be  determined 
with  certainty.  The  formulas  to  be  deduced  are  such  that,  in 
general,  they  give  limiting  maximum  values  under  the  different 
conditions,  and  the  hypothesis  here  adopted  has  the  practical 
advantage  of  erring  on  the.  side  of  safety  for  the  wall.  In  an 
unlimited  mass  of  earth  with  horizontal  surface,  the  pressure 
against  any  imaginary  vertical  plane  must  evidently  be  normal 
to  that  plane ;  now  if  a  wall  is  to  be  designed  to  replace  the 
earth  on  one  side,  the  pressure  against  its  back  will  also  be 
normal.  It  would  seem  then,  that  the  most  satisfactory  de- 
gree of  stability  of  the  earth  will  be  secured  by  designing  the 
wall  under  the  assumption  of  normal  pressure. 

The  views  just  expressed  are,  howevet,  not  accepted  by 
some  engineers  who  claim  that  the  actual  normal  pressure  is 
usually  less  than  the  values  theoretically  deduced  for  Pl ,  par 
ticularly  for  walls  that  have  been  observed  to  fail.  In  Article 
9  there  will  hence  be  investigated  formulas  for  the  pressure 
supposing  that  it  is  inclined  to  the  normal  to  the  back  of  the 
wall  at  an  angle  0;  the  value  of  the  pressure  thus  derived  will 
be  called  the  "  inclined  pressure,"  and  be  designated  by  Pv 


ART.  8.] 


NORMAL   PRESSURE  AGAINST    WALLS. 


Hence  either  Pl  or  P^  can  be  used  in  investigating  the  wall  as 
the  engineer  thinks  best. 

Problem  7.  Let  the  wall  in  Figure  7  be  vertical,  12  feet  in 
height,  its  thickness  uniformly  2  feet,  and  its  weight  3600 
pounds.  Let  the  point  of  application  of  P  be  4  feet  above 
the  base.  Compute  the  value  of  P  to  cause  rotation,  (a)  when 
the  angle  ft  is  o° ;  (b)  when  the  angle  ft  is  30°. 


ARTICLE  8.    NORMAJ.  PRESSURE  AGAINST  WALLS. 

In  Figure  8  is  shown  a  wall  which  sustains  the  lateral  press- 
ure of  a  bank  of  earth.  The  back  of  the  wall  BA  is  inclined  to 
the  horizontal  at  the  angle  6,  an^i  the  surface  of  the  earth.  AM 


FIG.  8. 


is  inclined  at  the  angle  8.  The  line  BCf  represents  the  natural 
slope  of  the  earth  with  the  inclination  <p.  It  is  required  to 
find  the  lateral  pressure  of  the  earth,  supposing  that  its  direc- 


28  THE  LATERAL  PRESSURE   OF  EARTH.       [CHAP.  II. 

tion  is  normal  to  the  back  of  the  wall,  or  that,  in  Figure  7, 
the  angle  2  is  zero. 

Draw  BM  making  any  angle  x  with  the  horizontal,  and 
consider  that  the  prism  ABM  in  attempting  to  slide  down  the 
plane  B M  exerts  a  pressure  upon  AB.  Let  IV  be  the  weight 
of  this  prism,  represented  by  the  line  OW,  and  let  it  be  re- 
solved into  a  component  P^  acting  normal  to  the  back  of  the 
wall,  and  a  component  R  acting  opposite  to  the  direction  of 
the  reaction  of  the  earth  below  BM.  Let  ON  be  normal  to 
the  plane  BM,  then  the  angle  NOR  will  be  equal  to  the 
angle  of  friction  of  earth  upon  earth,  if  the  prism  ABM  is  just 
on  the  point  of  sliding  down  BM\  (for,  as  ON  and  NR  are 
components  of  OR  the  triangle  gives  NR  =  ON.  tan  NOR, 
but  from  the  law  of  friction  NR=f.  ON\  hence /=  tan  0  = 
tan  NOR,  and  accordingly  NOR  =  0.) 

Now  in  the  triangle  IVOR  the  side  OW  represents  the 
weight  W,  and  the  side  WR  the  resultant  normal  pressure  P, . 
Hence 

p  -    TT/sm  WOR 
'""       sin  WRO' 


w  But  the  angle  WOR  is  x  —  0  and  the  angle  WRO  is  0  +  0  —  x. 
Let  h  be  the  vertical  height  of  the  wall,  and  w  the  weight  of  a 
cubic  unit  of  earth  ;  then  the  value  of  W  for  one  unit  in  length 
•of  the  wall  is 

,„,..       .     ,     sin  (0  —  #)  sin  (9  —  x) 
W  =    iv .  BA  .  BM.  sin  ABM  =  %wfr .v         /     -A__ — I 


-X^x  -         - 

; 


ART.  8.]  NORMAL  PRESSURE  AGAINST    WALLS.    '"'  ''  29 

TU    K       i     re 

The  above  value  of  Pl  then  can  be  written 

P  -  JL./«  sin  (g  ~  <*)  sin  (g  -  x)  sin  (^  -  0) 


which  expresses  the  normal  pressure  due  to  any  prism  whose 
plane  BM  makes  an  angle  x  with  the  horizontal.  This  expres- 
sion becomes  o,  both  when  x  =  6  and  when  x  =  0,  and 
between  those  limits  it  has  a  maximum  value  which  is  to  be 
taken  as  the  pressure  against  the  wall,  since  such  can  occur  if 
the  earth  is  about  to  slide  down  the  corresponding  plane. 

In    order   to    find    the    value    of  x  which  renders   (20)  a 
maximum  it  is  best  to  write  it  in  the  form 


in  which  y  =  cot  (8  —  x\  a  =  cot  (8  —  0),  b  =  cot  (8  —  <?), 
c  =  cot  (p,  A  —  ^w/f  sin  (0  —  0)  and  B  •=.  sin8  0  sin  0.  Dif- 
ferentiating this  with  respect  to  y  and  putting  the  first  deriva- 
tive equal  to  zero,  there  is  found 

y  =  a  +  )/(a  _  fi)  (J+Tj;     ....      (22) 
and  inserting  this  in  (21)  there  results  for  the  maximum 
p^  _A 

Thus  /\  is  expressed  in  terms  of  the  data  w,  k,8,d  and  0,  but 
to  obtain  a  more  convenient  form  it  is  well  to  replace  t-he 


3O  THE  LATERAL  PRESSURE   OF  EARTH.       [CHAP.  1L 

cotangents  by  their  equivalents  in  terms  of  sines.     Then  after 
reduction  it  becomes 


•P,= r=£2£fc£— «,    •    •(*!} 


sm 


which  is  the  formula  for  the  lateral  normal  pressure  of  a  bank 
of  earth  against  the  back  of  a  retaining  wall. 

This  formula  is  valid  for  any  value  of  0  greater  than  0,  and 
for  any  value  of  3  less  than  0.  By  its  discussion  simpler 
formulas  for  special  cases  can  be  deduced. 

The  greatest  value  of  d  will  be  that  of  the  natural  slope  0. 
For  this  case  formula  (23)  becomes 


(0  — 


which  is  the  greatest  normal  thrust  that  can  be  caused  by  a 
sloping  bank  ;  if  the  wall  be  vertical  6  =  90°  and  this  reduces 
to  the  simple  form  Pl  =  fywh*  cos2  0. 


The  most  common  case  is  that  where  the  surface  AM  Is 
horizontal  ;  for  this  6  =o  and  (23)  becomes 


<2  rt 



ART.  g.]          INCLINED   PRESSURE  AGAINST    WALLS.  31 

Avhich  is  the  normal  pressure  of  a  level  bank  of  earth  against 
an  inclined  wall.  If  in  this  6  =  90°,  there  results  the  formula 
for  the  pressure  of  a  level  bank  against  a  vertical  wall, 

tan'  (45  -  J0), 


which  is  the  well-known  expression  first  deduced  by  COULOMB 
in  1773. 

Problem  8.  Prove  from  (22)  that,  when  #  =  o,  the  plane 
BM  bisects  the  angle  between  BA  and  BC.  Prove  it  also  by 
making  d  =  o  in  (21),  and  then  equating  the  first  derivative  to 
.zero,  thus  deducing  x  =  £(#-(-  0). 


ARTICLE  9.    INCLINED  PRESSURE  AGAINST  WALLS. 

In  Figure  9  is  shown  a  wall  which  sustains  the  lateral  pres- 
sure of  a  bank  of  earth,  the  back  of  the  wall  being  inclined  to 
the  horizontal  at  an  angle  0,  and  its  vertical  height  being  h. 
The  upper  surface  of  the  earth  has  an  inclination  8  to  the 
horizontal,  which  is  not  greater  than  the  natural  slope  0.  It 
is  required  to  find  the  lateral  pressure  of  the  earth  supposing 
that  its  direction  makes  an  angle  0  with  the  normal  to  the 
back  of  the  wall. 

Draw  BM  making  any  angle  x  with  the  horizontal,  and 
consider  that  the  prism  BAM  in  attempting  to  slide  down 
this  plane  is  sustained  by  the  reactions  of  the  wall  and  of  the 
earth  below  BM.  Let  OW represent  the  weight  of  this  prism, 


THE  LATERAL   PRESSURE   OF  EARTH.        [CHAP.  II. 


and  let  it  be  resolved  into  components  OP^  and  OR  opposite 
in  direction  to  these  reactions.  Let  OL  be  normal  to  the 
back  of  the  wall,  and  ON  be  normal  to  the  plane  BM.  Then 
if  motion  is  just  about  to  occur,  the  angle  NOR  is  equal  to  the 
angle  of  friction  0  of  earth  upon  earth,  and  LOP9  is  equal  to 
the  angle  of  friction  0'  of  earth  upon  masonry.  Although  ft 


is  perhaps  in  general  greater  than  0,  it  is  customary  to  take  it 
as  the  same,  thus  erring  on  the  side  of  safety ;  accordingly 
LOP,  =  <f>. 

Let  Wbe  the  total  weight  of  the  earth  in  the  prism  BAM, 
and  w  its  weight  per  cubic  foot.  Let  P^  represent  the  inclined 
resultant  pressure  against  the  wall.  In  the  triangle  ROW,  the 
angle ROW'isx—  0,  and  ORWis0  +  20  -  x\  hence 


=  W 


sin  (x  —  0) 
sin(0{  +  20  —  x)  ' 


fu/-v 

ART.  9.]          INCLINED  PRESSURE  AGAINST    WALLS.  33 

The  weight  H^  for  one  unit  in  length  of  the  wall  is  w  X  area 
BAM  X  I.     The  area  of  BAM  equals  i^yi  .  BM  .  sin 
the  side  BA  is  /*  -r-  sin  0,  the  angle  y^J/  is  0  —  #,  and 


sm  y  sin     .  sin  *  — 

The   value   of   W  is   thus   expressed   in    terms  of  #  and  the 
given  data,  and  P9  becomes 

P  -  Iwtf  sin  (0  -  (?)  sin  (6  -  x)  sin  (x  -  0)  ,     . 

1  sin'  0  sin  (*-(?)  sin  (0  +  20  -  *)' 

which  gives  the  pressure  due  to  any  prism  BAM. 

The  greatest  possible  value  of  P2  is  to  be  regarded  as  the 
actual  value  of  the  inclined  pressure.  By  proceeding  as  in 
Article  8  it  can  be  shown  that  this  obtains  when 

cot  (6  -  x)  =  cot  (6  -  0) 

+  V  [cot  (0—  0)  —  cotT(0  —  £)]  [cot  (0  -  0)  +  cot  20]  >     (28) 

and  that  the  maximum  value  is 


^ 

•   a  /»   •    ru   i    -*         \        /    sin  2     .  sin       — 
sin2  6^  sin  (0  +  0)1  I  +  A  /   . 
;  y   sm 


lfi  ,     ,,    .    ,  . 

(0  -\-  0)  sin  (6  —  tf 


which  is  the  general  formula  for  the  so-called  inclined  pressure 
and  from  which  the  results  for  all  special  cases  can  be  deduced. 


34  THE  LATERAL  PRESSURE   OF  EARTH.       [CHAP.  II. 

The  greatest  slope  8  will  be  the  natural  slope  0.     For  this 
case  the  formula  reduces  to 


- 
3      sin2  B  sin  (0+0)' 


which  is  the  pressure  due  to  a  ba$k  of  maximum  slope  against 
an  inclined  wall.  If  the  back  of  the  wall  be  vertical,  0  =  90° 
and  the  expression  takes  the  simple  form  Pa  —  %wh*  cos  0. 


The  most  common  case  is  that  where  the  surface  AM  is 
horizontal  ;  for  this  d  =  o  and  (29)  becomes 


„ y~rf  sin2(<9—  0) 

/^sTn20sin0     y        (31) 
Vsin( 


sin' 


which  is  the  inclined  pressure  of  a  level  bank  of  earth  against 
an  inclined  wall.  If  in  this  6  —  90°,  there  results  the  formula 
for  a  level  bank  of  earth  retained  by  a  vertical  wall. 


(32) 


which  is  the  well-known  expression  deduced  by  PONCELET. 

Problem  9.  In  formula  (27)  make  0  =  90°  and  d  —  o°. 
Then  find  the  value  of  x  which  renders  it  a  maximum,  and 
deduce  the  corresponding  value  of  Pv 


ART.  10.]  FORMULA   FOR  LATERAL  PRESSURE.  35 


ARTICLE  10.    GENERAL  FORMULA  FOR  LATERAL 
PRESSURE. 

Let  a  wall  whose  back  is  AB  sustain  a  bank  of  earth  BAM 
as  in  Figure  9.  If  the  earth  be  loose,  the  weakest  plane  BM 
wil  be  that  along  which  rupture  is  about  to  occur,  so  that  the 
angle  NOR  =  0,  as  in  the  two  preceding  articles.  Let  the 
resultant  lateral  pressure  be  designated  by  g,  and  let  its  direc- 
tion make  an  angle  z  with  the  normal  to  the  wall  so  that 
LOP^  =  s.  By  the  same  reasoning  and  methods  as  before 
used,  it  is  found  that  the  expression  for  the  pressure  due  to 
any  prism  AB  M  and  the  value  of  cot  (d  —  x)  which  renders  it 
a  maximum  are  the  same  as  given  by  (27)  and  (28)  if  the 
single  term  20  be  replaced  by  0  -f-  z,  and  then  results 


p  =  _  _  f  (33) 

sin2  6  sin  (8  +  z](i  +  .  /  sin(0  +  *)  sin(0  -  tf)V  ' 
\         \/   sin(0  +  *)sin(0-tf)/ 


which  is  a  general  formula  for  the  lateral  pressure  in  terms  of 
the  unknown  angle  z.  If  z  =  o,  the  direction  of  P  is  normal 
to  the  back  of  the  wall  and  (33)  reduces  to  (23).  If  z  —  0,  the 
pressure  is  inclined  to  the  normal  at  the  angle  of  friction  and 
(33)  reduces  to  (28). 

From  the  above  formula  a  number  of  theories  of  earth 
pressure  can  be  deduced  by  making  different  assumptions  with 
regard  to  the  angle  z.  For  instance,  it  seems  to  some  authors 


36  THE  LATERAL   PRESSURE   OF  EARTH.       [CHAP.  IL 

a  reasonable  theory  which  makes  the  pressure  upon  a  vertical 
wall  parallel  to  the  earth  surface  A  M;  for  this  case  9  =  90°, 
and  z  =  90°  -\-  £  —  0,  and  inserting  these  in  (33)  it  reduces  to 


p  __  _          cos3  0        _ 

=  " 


cos 


which  is  RANKINE'S  formula  for  the  lateral  pressure  against  a 
vertical  wall.  In  like  manner  several  other  formulas,  more  or 
less  reasonable,  can  be  established.  But  probably  everything 
necessary  for  the  practical  engineer  is  given  in  Articles  8  and  9. 

Problem  10.  Deduce  formulas  for  the  earth  pressure 
under  the  supposition  that  its  direction  is  horizontal. 

ARTICLE  11.    COMPUTATION  OF  PRESSURES. 

In  computing  the  lateral  pressure  of  earth  from  the  above 
formulas  it  is  customary  to  take  h  in  feet  and  w  in  pounds  per 
cubic  foot ;  then  the  value  of  P  will  be  in  pounds  per  running 
foot  of  the  wall.  On  account  of  the  uncertainty  in  the  data 
the  trigonometric  functions  need  be  taken  only  to  three  or 
four  decimal  places,  or,  if  logarithms  be  used,  as  will  be  found 
most  convenient,  a  four-place  table  will  be  amply  sufficient. 
The  values  of  the  pressures  for  several  cases  will  now  be  com- 
pared, the  walls  all  being  18  feet  in  vertical  height,  the  earth 
weighing  100  pounds  per  cubic  foot  and  having  a  natural  slope 
0  =  34  degrees.  Here  the  value  of  ^.wh*  is  16200  pounds. 


ART.  ii.]  COMPUTATION  OF  PRESSURES.  37 

For  a  level  bank  of  earth  and  a  wall  whose  back  slopes 
backward  with  the  inclination  0  =  80°,  formulas  (25)  and  (31) 
give  the  pressures 

PI  =  3  57°»  ?*  =  2  590  pounds. 

For  a  level  bank  of  earth  and  the  back  of  the  wall  vertical,  for- 
mulas (26)  and  (32)  give 

Pl  =  4  580,  P3  —  4  210  pounds. 

For  a  level  bank  of  earth  and  the  back  of  the  wall  sloping 
forward  so  that  0  =  100°,  formulas  (25)  and  (31)  give 

Pl  =  5  760,  Pz  =  5  670  pounds. 

Here  it  must  be  remembered  that  the  direction  of  P  is 
normal  to  the  wall,  while  the  direction  of  Pt  makes  an  angle  of 
34°  with  the  normal  to  the  wall. 

For  the  same  walls  sustaining  earth  whose  upper  surface 
has  the  slope  #  =  10  degrees,  the  following  values  are  found 
from  formulas  (23)  and  (29): 

For  6  =  80°,  P,  =  3  920,  P2  =  3  400  pounds. 
For  0  =  90°,  Pl  =  5  080,  P2  —  4  960  pounds. 
For  8  =  100°,  P,  =  6  469,  P2  =  6  480  pounds. 

For  the  same  walls  sustaining  earth  whose  upper  surface 
has  the  angle  of  repose  3  =  34°,  formulas  (24)  and  (30)  give : 

For  6  —  80°,  P,  =  8  780,  P2  =  9  460  pounds. 
For  6  —  90°,  P,  =  ii  130,  Pz  =  13  430  pounds. 
For  0  —  100°,  P,  =  14  1 60,  P2  =  19  380  pounds. 


38  THE  LATERAL  PRESSURE   OF  EARTH.       [CHAP.  II. 

A  comparison  of  the  above  values  shows  that  the  pressure 
increases  both  with  0  and  d.  For  a  level  bank  of  earth  the 
values  of  P^  are  less  than  those  of  /J,  but  for  a  large  value  of 
d  the  values  of  P2  become  greater  than  those  of  Pr  Whether 
the  true  thrust  against  the  wall  is  Pl  or  P2 ,  or  some  inter- 
mediate value,  cannot  be  determined  theoretically,  and  hence 
the  best  procedure  for  the  engineer  will  be  to  use  those 
values  which  are  the  most  unfavorable  to  stability. 

For  the  case  of  water  <p  =  o  and  d  =  o,  and  all  the  for- 
mulas for  pressures  reduce  to 

p-  fyj?  +  sin  0, (35) 

in  which  w  is  the  weight  of  a  cubic  foot  of  water,  or  62% 
pounds.  The  pressure  of  water  against  a  wall  18  feet  in  verti- 
cal height  will  hence  be  10  280  pounds  when  0  =  80  degrees, 
10  125  pounds  when  0  =  90  degrees,  and  10  280  pounds  when 
6  =  100  degrees,  its  direction  being  always  normal  to  the  back 
of  the  wall. 

Problem  n.  Compute  the  pressures  against  a  wall  9  feet 
in  vertical  height  for  earth  weighing  100  pounds  per  cubic  foot 
and  having  an  angle  of  repose  0  =  34  degrees,  (a)  for  the  case 
when  #  =  10  degrees  and  6  =  80  degrees  ;  (b)  for  the  case  when 
d  =  20  degrees  and  6  =  80  degrees. 


ART.  12.]  THE   CENTRE   OF  PRESSURE.  39 


ARTICLE  12.    THE  CENTRE  OF  PRESSURE. 

For  all  the  above  cases  the  formulas   for   the   resultant 
lateral  pressure  of  the  earth  may  be  written 

P  =  \wtf  .  k, 

in  which  k  is  a  function  of  the  angles  #,  $,  and  0.  If  y  repre- 
sent any  vertical  depth  measured  downward  from  the  top,  the 
resultant  pressure  on  the  part  of  the  wall  corresponding  to 
this  depth  is 


which  shows  that  the  resultant  pressure  varies  as  the  square  of 
the  height  of  the  wall.  The  pressure  per  square  unit  at  any 
point  on  the  wall  varies,  however,  directly  as  the  height,  for 
if  y  be  increased  by  dy  the  increase  in  P  is  dP  or  wkydy,  and 
the  pressure  per  square  unit  over  the  area  I  X  dy  is 


dp 


The  laws  governing  the  distribution  of  earth  pressures  are 
hence  the  same  as  for  water,  the  unit-pressure  at  any  point 
varying  as  the  depth,  and  the  total  pressure  as  the  square  of 
the  depth. 


4O  THE   LATERAL   PRESSURE    OF  EARTH.        [CHAP.  II. 

The  point  where  the  resultant  pressure  P  is  applied  to  the 
wall  is  called  the  centre  of  pressure,  and  this  is  at  a  vertical 
distance  from  the  top  of  the  wall  equal  to  two-thirds  its  height. 
This  may  be  proved  by  Figure  10,  which  gives  a  graphical 
representation  of  the  pressure  against  the  back  of  the  wall,  the 
unit-pressures  falling  into  a  triangular-  shape,  since  each  is 
proportional  to  its  distance  below  the  top.  The  point  of 


FIG.  10. 

application  of  the  resultant  pressure  P  hence  passes  through 
the  centre  of  gravity  of  this  triangle  and  cuts  the  back  at 
two-thirds  its  length  from  the  top. 

For  another  proof  the  principle  of  moments  may  be  used. 
Let  y  beany  vertical  distance  from  the  top,  and  y'  the  vertical 
distance  from  the  top  to  the  centre  of  pressure.  Then  taking 
the  top  of  the  back  of  the  wall  as  the  origin  of  moments, 


Inserting  in   this  the   values  of  P  and  dP  given  above  and 

Integrating  betwen  the  limits  y  —  o  and  y  =  //,  there  results 

x 

/  =  **;  ........    (36) 


ART.  12.]  THE   CENTRE    OF  PRESSURE.  41 

that  is,  the  centre  of  pressure  is  at  a  vertical  distance  \h  below 
the  top  of  the  wall,  or  at  \h  above  the  base. 

Problem  12.  Let  a  level  bank  of  earth  have  a  load  of  q 
pounds  per  square  foot  upon  the  surface  AM.  Show  that  the 
resultant  normal  pressure  due  to  both  bank  and  load  is 


and  that  the  depth  of  the  centre  of  pressure  below  the  top  of 
the  wall  is 

_  2wh  +  3?       ,  (      } 

- 


Find  the  position  of  the  centre  of  pressure  when  h  =  18  feet, 
<w  =  100  pounds  per  cubic  foot,  and  q  =  300  pounds  per 
square  foot. 


INVESTIGATION  OF  RETAINING    WALLS.  [CHAP.  III. 


CHAPTER   III. 


INVESTIGATION  OF  RETAINING  WALLS. 


ARTICLE  13.    WEIGHT  AND  FRICTION  OF  STONE. 

The  lateral  pressure  of  the  earth  against  a  retaining  wall 
tends  to  cause  failure  in  two  ways,  namely,  by  sliding  and  by 
rotation.  This  tendency  is  resisted  by  the  friction  between 
the  stones  and  by  the  weight  of  the  wall.  The  following  table 
gives  average  values  of  the  unit-weights  and  the  coefficients  of 
friction  for  different  kinds  of  masonry: 


Kind  of  Masonry. 

Coefficient 
of 
Friction. 

Angle 
of 
Friction. 

Weight. 

Pounds  per 
cubic  foot. 

Kilos  per 
cubic  meter. 

Limestone  and  Granite  : 
Ashlar  Masonry  

0.6 

0.6 
0.65 

31° 

31° 

33° 

165 
150 
125 

150 
130 
no 

100 

2640 
2400 
2000 

2400 
2100 
1760 
1600 

Large  Mortar  Rubble..  .  . 
Small  Dry  Rubble    .    . 

Sandstone  : 
Ashlar  Masonry             . 

Large  Mortar  Rubble  
Small  Dry  Rubble  

Coarse  Brickwork     .   .    .  . 

ART.  13.]  WEIGHT  AND   FRICTION  OF   STONE.  43 

In  the  investigation  of  masonry  walls  the  influence  of  the 
mortar  is  generally  neglected,  on  account  of  its  uncertain 
character  and  because  the  error  is  then  on  the  side  of  safety. 
The  above  coefficients  of  friction  are  hence  stated  for  dry 
masonry,  and  will  probably  be  somewhat  increased  when  mor- 
tar is  used.  For  rubble  masonry  the  coefficient  of  friction  is 
often  somewhat  higher  than  for  ashlar ;  but  its  value  is  so  un- 
certain that  no  figure  is  given  in  the  table. 


The  coefficient  of  friction  of  stone  upon  stone  is  deter- 
mined by  placing  two  plane  surfaces  together  and  then 
gradually  inclining  the  surface  of  contact  until  the  upper 
stone  begins  to  slide  upon  the  lower.  The  angle  made  by  the 
plane  with  the  horizontal  is  the  angle  of  friction,  and  its  tan- 
gent is  the  coefficient  of  friction,  as  shown  by  equation  (2). 

The  word  "  batter  "  means  the  inclination  of  the  face  or 
back  of  a  wall,  measured  by  the  ratio  of  its  horizontal  to  its 
vertical  projection,  or  in  inches  of  horizontal  projection  per 
foot  of  vertical  height.  Let  0  be  the  angle  of  inclination  of 
the  back  of  the  wall  to  the  horizontal,  as  in  Figure  8.  Then 
cot  6  is  the  batter  of  the  back,  and  the  values  of  0,  sin  8, 
and  cos  0  for  different  batters  are  given  in  the  following  table. 
If  the  back  of  the  wall  leans  backward,  0  is  less  than  90  de- 
grees, and  cos  0  and  cot  0  are  positive  ;  if  it  leans  forward, 
6  is  greater  than  90  degrees  and  cos  0  and  cot  0  are  negative  ; 
sin  0  is  positive  in  both  cases.  These  values  will  be  useful 
in  many  computations. 


INVESTIGATION  OF  RETAINING    WALLS.  [CHAP.  Ill, 


Batter  in 
inches 
per  foot. 

Angle  &  for 
Backward 
Batter. 

Angle  0 
for 
Forward  Batter. 

sin  0. 

cos  6. 

Batter 
cot  6. 

O 

90°  oo' 

90°  oo' 

I.OOOO 

O.OOOO 

0.0000 

i 

87    37 

92    23 

0.9991 

0.0461 

0.0417 

I 

85    14 

94    46 

0.9965 

O.C83I 

0.0833 

Ii 

82    52 

97    08 

0.9923 

0.1239 

o.  1250 

2 

80   32 

99    28 

0.9864 

0.1645 

0.1667 

2* 

78    14 

101    46 

0.9790 

0.2039 

0.2083 

3 

75    58 

104    02 

0.9702 

0.2425 

O.25OO 

3* 

73   45 

106    15 

o  .  9600 

0.2797 

0.2916 

4 

71    34 

108    26 

0.9487 

0.3162 

0-3333 

5 

67   29 

112     31 

0.9239 

0.3828 

0.4144 

6 

63    26 

116   34 

0.8944 

0.4472 

O.5OOO 

Problem  13.  Compute  the  values  of  sin  6,  cos  8,  tan  0,  and 
cot  6  for  a  batter  of  4^  inches  per  foot ;  also  for  a  batter  of  5 •£ 
inches  per  foot. 


ARTICLE  14.    GENERAL  CONDITIONS  REGARDING  SLIDING. 

A  retaining  wall  may  fail  by  sliding  on  its  base  or  on  some 
joint  above  the  base.  When  a  wall  is  just  on  the  point  of 
failure  it  is  in  the  state  of  equilibrium,  that  is,  the  weight  of  the 
wall  just  balances  the  pressure  and  reaction  of  the  earth.  The 
proper  state  of  a  wall  is,  of  course,  stability  ;  and  failure  brings 
disgrace  upon  its  designer. 

The  degree  of  stability  of  a  wall  against  sliding  may  be  in- 
dicated by  a  number  called  the  factor  of  security  which  ranges 
in  value  from  unity  to  infinity.  This  factor  will  be  designated 
iby  n ;  when  n  =  i  equilibrium  exists  and  the  wall  will  fail , 


ART.  14.]  GENERAL    CONDITIONS  REGARDING  SLIDING.  45 

when  n  >  I  the  wall  is  stable  and  its  degree  of  stability  varies 
with  n  ;  when  n  =  oo  the  highest  possible  state  of  stability 
exists. 

The  analytical  conditions  of  equilibrium  and  stability  for 
the  case  of  sliding  are  the  following.  Let  Figure  II  represent 
two  bodies  having  a  plane  surface  of  contact,  N  the  total  force 


normal  to  that  plane,  F  the  total  force  parallel  to  it,  and  /the 
coefficient  of  friction  between  the  surfaces.  Then  the  condi- 
tion of  equilibrium  is,  as  in  (i), 

F  =  fN,     .......    (39) 

and  the  condition  of  stability  is 

F<fN        or        nF  =  ftf,    ....     (40) 


in  which  n  is  a  number  greater  than  unity  called  the  factor  of 
security.  The  equation  (40)  may  be  used  for  the  discussion  of 
all  cases  of  sliding,  .Fand  N  being  the  sum  of  the  components 


46  INVESTIGATION  OF  RETAINING    WALLS.  [CHAP.  IIL 

in  the  directions  parallel  and  normal  to  the  plane  of  all  the 
forces  exerted  by  one  body  on  another.  If  R  be  the  resultant 
of  all  these  forces  and  C  be  the  angle  which  it  makes  with  the 
normal  ON,  the  value  of  F  is  R  sin  C,  and  that  of  N  is  R  cos  C- 
Inserting  this  in  (40)  it  becomes 


n  tan  £  =/,.".     .....     (41) 

which  is  another  form  of  the   condition  of  stability  against 
sliding. 

The  graphical  conditions  of  equilibrium  and  stability  for 
sliding  are  simple.  In  Figure  1  1  let  ON  be  normal  to  the 
plane  of  contact  and  NOF  be  the  angle  of  friction  0,  that 
is,  the  angle  whose  tangent  is  f.  Let  R  make  an  angle  C  with 
the  normal  ON,  then  equilibrium  obtains  when  C  equals  0, 
and  stability  occurs  if  C  is  less  than  0.  Draw  NF  parallel  to 
the  plane  of  contact,  and  let  T  be  the  point  where  it  inter- 
sects the  line  of  direction  of  R.  The  position  of  T  indicates 
the  degree  of  stability  against  sliding  ;  if  the  distances  NT 
and  NF  be  determined,  the  factor  of  security  is  the  ratio  of 
the  latter  to  the  former,  or 


NF 

(42> 


for,  it  is  seen  that  this  formula  is  the  same  as  (41),  NF  being/, 
and  NT  being  tan  C  if  the  distance  ON  be  unity,  and  their 
ratio  being  the  same  as  these  tangents  whatever  be  the  length 
of  ON. 


ART.  15.]        '  GRAPHICAL  DISCUSSION  OF  SLIDING.  47 

Problem  14.  A  plane  surface  is  inclined  at  an  angle  of  40° 
to  the  horizontal,  and  on  it  is  a  block  weighing  125  pounds, 
against  which,  to  prevent  it  from  sliding,  a  horizontal  force 
of  300  pounds  acts.  If  the  angle  of  friction  of  the  block  upon 
the  plane  is  18°,  compute  the  factor  of  security  against  sliding. 


ARTICLE  15.    GRAPHICAL  DISCUSSION  OF  SLIDING. 

Let  Figure  12  represent  the  section  of  a  wall  whose  dimen- 
sions and  weight  are  given.  Let  BC  be  any  joint  extending 
through  the  wall,  and  let  P  be  the  lateral  pressure  of  the  earth 
above  B.  It  is  required  to  investigate  the  security  of  the  wall 
against  sliding. 

The  pressure  Pis  applied  on  the  back  of  the  wall  at  one 
third  of  its  height  above  B,  and  its  direction  depends  on  the 
hypothesis  adopted  in  its  computation  ;  if  Article  8  is  used,  it  is 
normal  to  the  back  of  the  wall ;  if  Article  9,  it  is  inclined  at  an 
angle  equal  to  the  angle  of  natural  slope  of  the  earth. 

A  drawing  of  the  given  cross-section  is  made  to  scale,  and 
its  centre  of  gravity  found  :  this  is  G  in  the  figure.  The  area 
of  this  cross-section  is  then  determined  and  called  A  ;  if  this 
be  multiplied  by  ^,  the  weight  of  a  cubic  unit  of  masonry, 
the  product  is  V,  the  weight  of  a  wall  one  unit  in  length,  or 
V=  vA. 

Through  G  a  vertical  line  is  drawn,  and  the  direction  of  P 
is  produced  to  intersect  this  in  O.  Lay  off  OP  to  scale  equal 


48 


INVESTIGATION  OF  RETAINING    WALLS:  [CHAP.  III. 


to  the  earth  pressure  P,  and  OV  equal  to  the  weight  of  the 
wall,  V.  Complete  the  parallelogram  of  forces  OPRV,  thus 
finding  OR  as  the  resultant  of  P  and  V. 

Produce  OR  to  meet  the  joint  BC  in  T.  Through  O  draw 
ON  normal  to  BC,  and  then  draw  OF,  making  the  angle  NOF 
equal  to  the  angle  of  friction  of  stone  upon  stone.  This  com- 
pletes the  graphical  work. 


If  the  point  T  falls  between  N  and  F,  the  wall  will  not  fail 
by  sliding,  and  its  stability  is  the  greater  the  nearer  T  is  to  N* 
If  ^coincides  with  F9  the  wall  is  just  on  the  point  of  sliding 
along  the  joint  BC,  and  much  more  so  is  this  the  case  if  T 
falls  beyond  F.  As  explained  in  the  last  Article,  a  numerical 
expression  of  the  degree  of  stability  can  be  obtained  by  divid- 
ing the  distance  NF  by  NT,  or  if  n  be  the  factor  of  security 
against  sliding, 

_NF 

n~  NT 


ART.  15.]  GRAPHICAL  DISCUSSION  OF  SLIDING.  49 

This  becomes  unity  when  NT  equals  NF,  and  infinity  when 
NT  is  zero,  the  first  value  indicating  the  failure  of  the  wall 
and  the  second  the  greatest  possible  degree  of  stability  against 
sliding.,  It  is  recommended  that  for  first-class  work  n  should 
not  be  less  than  3.0,  and  fortunately  it  is  always  easy  in  build- 
ing a  wall  to  make  its  value  greater  than  this  by  properly  in- 
clining the  joints  (Article  23). 

The  above  method  applies  either  to  the  base  of  the  wall  or 
to  any  joint  that  extends  through  it,  whether  the  joint  be  hori- 
zontal or  inclined.  Owing  to  the  uncertainty  regarding  the 
weight  and  angle  of  repose  of  the  earth,  the  direction  of  P,  and 
the  angle  of  friction  of  the  stone,  it  will  not  always  be  possible 
to  obtain  values  of  the  factor  of  security  which  are  perfectly 
satisfactory.  Still  the  investigation  will  generally  determine 
if  danger  exists,  and  of  course  unfavorable  values  of  the  data 
should  be  used  in  the  analysis.  If  the  wall  have  no  joints  ex- 
tending through  it,  an  anafysis  for  sliding  need  not  be  made. 

Problem  14.  Prove  that  the  centre  of  gravity  of  a  quadri- 
lateral abed  can  be  found  as  follows  :  Draw  the  diagonal  ac  and 
bisect  it  in  e ;  join  be,  and  take  ef  equal  to  \be ;  through  / 
draw  fk  parallel  to  bd.  Draw  the  other  diagonal  bd  and  bisect 
it  in  g;  join  g<?rand  take  gh  equal  to|%;  through  g  draw  gk 
parallel  to  ac.  The  centre  of  gravity  is  at  k,  the  intersection 
of  fk 


INVESTIGATION  OF  RETAINING    WALLS.  [CHAP,  III. 


ARTICLE  16.    ANALYTICAL  DISCUSSION  OF  SLIDING.    * 

Let  A  BCD  represent  the  cross-section  of  a  wall  whose 
dimensions  and  weight  are  given,  0  being  the  inclination  of 
its  back  to  the  horizontal.  Let  BC  be  any  joint  extending 
through  the  wall,  and  a  its  inclination  to  the  horizontal.  Let 
P  be  the  lateral  pressure  of  the  earth  above  this  joint,  and  2 
the  angle  between  its  direction  and  the  normal  to  the  wall ; 


=Z2%gF^ 

FIG.  13. 

if  Pbe  the  pressure  computed  by  the  formulas  in  Article  8,  the 
value  of  z  is  simply  zero  ;  if  by  those  in  Article  9,  its  value 
is  the  angle  of  repose  of  the  earth  ;  if  z  be  assumed  at  any 
intermediate  value,  P  is  computed  from  (33).  Let  V  be  the 
weight  of  the  wall.  It  is  required  to  investigate  the  degree  of 
stability  against  sliding. 

Let  .Fand  N  be  the  sum   of  the  components  of  P  and  V 
respectively  parallel  and  normal  to  the  joint,  and  f  the  coeffi- 


ART.  1  6.]        ANALYTICAL  DISCUSSION  OF  SLIDING.  51 

cient  of  friction.     Then  if  n  be  the  factor  of  security,  nF  =  fN, 
and 


(43) 


Now  by  resolving  P  and  V  parallel  and  normal  to  the  joint 
there  is  found 


F=  Psm(6+a  +  z)—  Fsin  a, 
N=  F  cos  or  -/>cos(#  +  <*  +  £) 


.and  if  these  be  inserted  in  (43),  the  value  of  n  is  expressed  in 
terms  of  the  given  data.     The  entire  analytical  discussion  of  '. 
the  sliding  of  a  wall  along  a  joint  consists  in  computing  n  from 
these  formulas.     If  n  is  greater  than  3,  the  security  against  ; 
sliding  is  ample  ;  if  n  is  less  than  3,  the  wall  does  not  have 
proper    security   for   first-class   work;    if    n  =  I,    failure   will 
occur. 

For  example,  consider  a  sandstone  wall  18  feet  high,  3  feet 
wide  at  the  top  and  6  feet  wide  at  the  base,  the  back  being 
vertical.  The  weight  of  the  masonry  is  taken  at  140  pounds 
per  cubic  foot,  and  the  coefficient  of  friction  on  the  horizontal 
joint  at  the  base  is  0.5.  This  wall  supports  a  level  bank  of 
earth  weighing  100  pounds  per  cubic  foot  and  having  an  angle 
of  natural  slope  of  34  degrees.  It  is  required  to  find  its  factor 
of  security  against  sliding. 

First,  let  the  pressure  P  and  its  direction  be  taken  from 
Article  8.  Here  h  =  18  feet,  w  =•-  100,  #  =  90°,  0=34°, 


52  INVESTIGATION   OF  RETAINING    WALLS.   [CHAP,  III. 

\  1^ 

d  =  o°,  and  z  —  o°.     Then  by  formula  (26)  there  is  computed 

•j 
P  =  4580  pounds.     The  weigut  of  the  wall  is 

V=  140  X  1 8  X  44  =  11340  pounds. 

Now  since  a  =  o°,  F=  4580  and  ^V  =  11340,  hence  the  factor 
of  security  is 

11340  X  0.5 


which  indicates  a  ver£  low  degree  of  stability. 

Secondly,  let  the  pressure  P  and  its  direction  be  taken  from 
Article  9.  Here  z  =  34°,  and  using  formula  (32)  there  is  found 
P=  4210  pounds.  Fis  11340  pounds  as  before.  From  (44), 

F  =  4210  sin  (90°  +  34°)  =  3490, 

N=  1 1340  —  4210  cos  (90°  +  34°)  =  13690, 

and  then  from  (43)  there  results  the  factor 

_  0.5  X  1 3690  _ 
3490 

which  indicates  a  degree  of  stability  too  low  for  first-class 
work. 

Unfortunate  indeed  it  is  that  the  theory  of  earth  pressure 
is  not  sufficiently  explicit  to  determine  the  exact  value  and 


ART.  16.]        ANALYTICAL  DISCUSSION  OF  SLIDING.  53 

direction  of  P.  He  who  believes  the  theory  of  Article  8  must 
conclude  that  this  wall  is  in  a  very  dangerous  condition  and 
almost  about  to  slide ;  he  who  defends  the  theory  of  Article  9 
might  conclude  that  it  is  not  in  great  danger,  and  that  its 
degree  of  security  is  fair.  It  is  well,  however,  not  to  forget 
that  the  given  data  are  liable  to  variations  fully  as  serious  as 
the  defects  in  the  theory.  Imagine  a  heavy  rainfall  to  increase 
w  and  decrease  0 ;  this  causes  P  to  become  larger,  and  as  F 
usually  would  be  smaller  in  wet  weather,  it  is  seen  that  the 
degree  of  stability  of  the  wall  would  be  greatly  diminished. 
If  the  factor  of  security  be  computed  for  both  theories  as  is 
done  above,  and  the  variation  in  the  data  be  regarded,  a  fair 
conclusion  can  generally  be  made  regarding  the  security  of  the 
wall.  The  effect  of  the  variable  data,  however,  is  often  so 
great  that  a  ripe  judgment,  based  upon  experience,  may  be 
more  reliable  than  computations. 

Problem  16.  Owing  to  a  heavy  rainfall  the  earth  behind 
the  above  wall  is  increased  in  weight  to  120  pounds  per  cubic 
foot  and  the  angle  of  natural  slope  is  decreased  to  32  degrees, 
while  the  coefficient  of  friction  at  the  base  of  the  wall  becomes 
0.45.  Compute  the  factor  of  security  of  the  wall  against 
sliding,  (a)  using  the  theory  of  Article  8,  and  (b)  using  that  of 
Article  o. 


54 


INVESTIGATION  OF  RETAINING    WALLS.  [CHAP.  III. 


ARTICLE  17.  GENERAL  CONDITIONS  REGARDING  ROTATION. 

Let  Figure  14  represent  two  bodies  having  the  plane  of 
contact  BC.  Let  M  be  the  middle  point  of  BC.  Let  R  be 
the  resultant  of  all  the  forces  which  each  body  exerts  on  the 
other,  and  let  T  be  its  point  of  application  on  BC.  It  is 
clear  that  rotation  or  overturning  will  instantly  occur  if  T  falls 
without  BC,  that  equilibrium  obtains  if  T  coincides  with  C, 


D 


FIG. 


and  that  stability,  more  or  less  secure,  will  result  if  T  falls 
within  BC.  The  nearer  the  point  of  application  T  is  to  the 
middle  of  the  base  M  the  greater  is  the  degree  of  stability 
against  rotation. 

To  investigate  the  degree  of  security  of  a  given  wall 
against  rotation  it  is  only  necessary  to  find  the  distance  MT 
either  graphically  or  analytically.  Let  n  be  the  factor  of 
security  of  the  wall,  then 


n  = 


MC_ 
MT 


(45) 


ART.  17.]  CONDITIONS  REGARDING  ROTATION.  5$ 

If  MT  equals  MC,  the  value  of  n  is  unity  and  failure  by  rota, 
tion  is  about  to  occur ;  if  MT  is  less  than  MC,  the  value  of  n 
is  greater  than  unity  and  the  wall  is  more  or  less  stable ;  if 
MT  is  zero,  n  is  infinity  and  the  wall  has  the  greatest  possible 
degree  of  stability. 

The  factor  of  security  n  should  not  have  a  value  less  than 
three  for  proper  stability.  To  demonstrate  this,  consider  the 
distribution  of  pressures  in  a  joint  as  represented  in  Figure  15. 


In  the  first  diagram  the  resultant  pressure  R  is  applied  at  the 
middle  of  BC\  here  the  pressure  will  be  uniformly  distributed 
over  the  joint,  and  the  unit-stress  5,  at  B  will  be  equal  to  the 
unit-stress  5  at  C.  In  the  second  diagram  the  resultant  R  is 
applied  so  that  MT  has  a  small  value ;  then  the  pressure  is 


56  INVESTIGATION  OF  RETAINING    WALLS.  [CHAP.  III. 

not  uniformly  distributed  over  the  joint,  but  the  unit-stress 
St  at  B  becomes  smaller  than  in  the  first  diagram,  while  the 
unit-stress  S  at  C  becomes  greater,  and  the  unit-stresses  be- 
tween B  and  C  are  taken  as  varying  proportionally.  In  the 
third  diagram  the  distance  MT  is  such  that  the  unit-stress  at 
B  becomes  zero ;  this  occurs  when  CT  is  one-third  of  CB 
(since  the  line  of  direction  of  R  passes  through  the  centre 
of  gravity  of  the  stress  triangle)  or  when  MT  is  one-third  of 
MC*  In  the  last  diagram  MT  has  become  greater  than  one- 
third  of  MC,  so  that  the  pressure  is  only  distributed  over 
CB'  and  the  portion  BB'  is  either  brought  into  tension  or  the 
joint  opens.  As  masonry  joints  cannot  take  tension  this  last 
is  a  dangerous  condition.  Therefore  the  ratio  of  MC  to  MT, 
or  the  factor  of  security,  should  not  be  less  than  3.0. 

If  the  joint  BC  be  divided  into  three  equal  parts,  so  that 
BD  =  DE  =  EC,  the  portion  DE  is  called  the  "  middle  third," 
and  the  above  requirement  is  otherwise  expressed  by  saying 
that  for  proper  security  against  rotation  the  resultant  of  all  the 
forces  above  any  joint  must  be  within  the  middle  third  of  that 
joint. 

Problem  17.  In  Figure  15  let  BC  be  horizontal,  and  let 
ABCD  be  a  cubical  block  weighing  625  pounds.  Compute  the 
factor  of  security  against  rotation  when  a  horizontal  force  of 
250  pounds  is  applied  at  A. 


ART.  18.]         GRAPHICAL  DISCUSSION  OF  ROTATION.  $? 


ARTICLE  18.    GRAPHICAL  DISCUSSION  OF  ROTATION. 

Let  Figure  12  represent  the  cross-section  of  a  wall  whose 
dimensions  and  weight  are  given.  Let  BC  be  any  joint  ex- 
tending through  the  wall,  and  let  P  be  the  lateral  pressure  of 
the  earth  above  B.  It  is  required  to  investigate  the  security 
of  the  wall  against  rotation. 

The  pressure  P  is  applied  on  the  back  of  the  wall  at  one- 
third  of  its  height  above  B  (Article  12),  and  its  direction  is 
•either  normal  to  the  wall  (Article  8),  inclined  to  the  normal  at 
the  angle  of  natural  slope  of  the  earth  (Article  9),  or  it  has  a 
direction  between  these  two  limits  (Article  10). 

A  drawing  of  the  given  cross-section  is  made  to  scale,  and 
its  centre  of  gravity  found  ;  this  is  at  G.  The  area  of  this  cross- 
section  is  next  determined  and  called  A  ;  then  the  weight  of 
the  wall  for  one  unit  in  length  is  V  =•  vAy  where  v  is  the  weight 
of  the  masonry  per  cubic  unit. 

Through  G  draw  a  vertical  line  and  produce  P  to  intersect 
it  in  O.  Lay  off  OP  to  scale  equal  to  the  earth  pressure  P, 
and  OV  equal  to  the  weight  V.  Complete  the  parallelogram 
of  forces  OPRV,  thus  finding  OR  as  the  resultant  of  P  and  V. 
Produce  OR  to  meet  the  joint  BC  in  71  Mark  M  as  the 
middle  point  of  BC,  and  measure  MT  and  MC.  This  com- 
pletes the  graphical  work. 


58  INVESTIGATION  OF  RETAINING    WALLS.  [CHAP.  III. 

If  T  falls  at  C,  the  wall  is  on  the  point  of  rotation  ;  and  if 
at  My  it  has  the  highest  possible  degree  of  stability.  If  BC 
be  divided  into  three  equal  parts  and  T  is  found  within  the 
middle  one,  the  wall  has  proper  security  against  rotation.  If 
it  falls  without  the  middle  third,  it  is  deficient  in  security 
(Article  17).  Dividing  MC  by  MT  the  factor  of  security  is 
found,  or 


If  this  is  unity  or  less,  the  wall  fails ;  if  it  be  smaller  than  3,  the 
wall  is  stable  but  not  secure ;  if  it  be  greater  than  3,  the  degree 
of  security  is  sufficient  as  far  as  rotation  alone  is  concerned  ;  if 
it  be  infinity,  nothing  more  can  be  desired. 

By  this  method  but  one  construction  is  needed  for  the 
investigation  of  a  wall  against  both  sliding  and  rotation.  It 
will  usually  be  found  for  ordinary  cases  that  the  factor  of 
security  against  rotation  is  least  for  the  base  of  the  wall  or  for 
the  lowest  joint.  For  the  general  discussion  the  base  of  the 
wall  is  drawn  inclined  in  Figure  12,  but  in  the  actual  drawing 
it  will  be  best  to  take  it  as  horizontal. 

Problem  1 8.  Let  a  wall  with  vertical  .back  support  a  level 
bank  of  sand  weighing  100  pounds  per  cubic  foot  and  having 
an  angle  of  repose  of  34  degrees.  Let  the  top  of  the  wall  be 
2  feet  thick,  its  base  7.57  feet,  its  vertical  height  20  feet,  and 
its  weight  per  cubic  foot  165  pounds.  Determine  the  factors 
of  stability  against  sliding  and  rotation  for  the  horizontal 
base,  taking  the  earth  pressure  from  Article  8.- 


ART.  19.]     ANALYTICAL  DISCUSSION  OF  ROTATION. 


59 


ARTICLE  19.    ANALYTICAL  DISCUSSION  OF  ROTATION. 

Let  ABCD  be  a  cross-section  of  a  wall  with  vertical  back, 
AB  being  24  feet,  the  top  AD  being  3  feet  and  the  base  BC 
being  8  feet.  Let  the  weight  per  cubic  foot  of  the  masonry 


be  150  pounds,  and  let  it  be  required  to  determine  the  factor 
of  security  against  rotation  for  a  horizontal  earth  pressure  P 
of  4000  pounds. 

Let  T  be  the  point  where  the  resultant  pierces  the  base, 
and  let  CT  be  represented  by  t ;  then  the  factor  of  security  is 

MC_      4 
MT         -t* 


60  INVESTIGATION  OF  RETAINING    WALLS.  [CHAP.  III. 

in  which  t  is  to  be  determined.  To  do  this,  drop  Dd  perpen- 
dicular to  BC,  dividing  the  cross-section  into  a  rectangle  of 
weight  Vl  and  a  triangle  of  weight  F3.  The  value  of  Vl  is 
15°  X  3  X  24  or  10,800  pounds,  and  its  horizontal  distance 
from  T  is  (6£  —  t)  feet.  The  value  of  V  is  150  X  5  X  12  or 
9000  pounds,  and  its  horizontal  distance  from  T  is  (f  X  5  —  /) 
feet.  The  lever  arm  of  P  with  reference  to  T  is  8  feet,  and  as 
R  passes  through  T  its  lever  arm  is  o.  Then  the  equation  of 
moments  is 

8000  X  8  =z  io8oo(6J  —  t)  +  9000(3^  —  /), 

from  which  /  is  found  to  be  1.83  feet,  and  then  the  factor  of 
security  against  rotation  is 

*  =  IT;  =  '•» 

which  is  not  sufficient  for  proper  stability. 

A  general  discussion  applicable  to  any  trapezoidal  cross- 
section  will  now  be  given.  Let  h  be  the  vertical  height  of  the 
wall,  a  the  thickness  of  the  top  AD,  b  the  thickness  of  its  base 
BC,  and  v  its  weight  per  cubic  foot.  Let  0  be  the  angle 
which  the  back  of  the  wall  makes  with  the  horizontal,  and  z  the 
angle  which  the  earth  pressure  P  makes  with  the  normal  to 
the  back  of  the  wall.  The  point  of  application  of  P  is  at  a 
vertical  height  of  \  h  above  B. 

Let  V  be  the  weight  of  the  wall  acting  through  the  centre 
of  gravity  of  the  cross-section,  and  let  5  be  the  point  where  its 
direction  cuts  the  base.  Let  R  be  the  resultant  of  P  and  V 


ART.  19.]     ANALYTICAL  DISCUSSION  OF  ROTATION. 


61 


acting  at  some  point  T  on  the  base.  Let  s  represent  the  dis- 
tance ££„  and  t  the  distance  CT.  Let  the  point  T  be  taken 
as  a  centre  of  moments,  and  let  the  lever-arm  of  P  with 


N 


FIG.  17. 


reference  to  it  be  /.      The  lever-arm  of   V  is  b  —  s  —  t,  and 
that  of  R  is  zero.     Then  the  equation  of  moments  is 


Pp=V(b-s-  t\ 


(46) 


which  is  the  fundamental  formula  for  the  investigation  of  re- 
taining walls.     This  may  be  written 


which  is  sometimes  a  more  convenient  form  for  use,  since  Vb 
and  Vs  can  be  treated  like  single  quantities. 

To  investigate  a  wall,  the  factor  of  security  n  is  to  be  de- 
termined.    From  formula  (45), 


n  = 


MC 


MT  '    \b  - 


(48) 


^x 

62  INVESTIGATION  OF  RETAINING    WALLS.  [CHAP.  IIL 

and  n  will  be  known  as  soon  as  /  is  found.     To  do  this  the 
value  of/  is  expressed  in  terms  of  /,  thus  : 


and  this  being  inserted  in  (47)  there  is  deduced 

•  /tf  cos(0  +  *) 


sn 


In  this  formula  F  is  the  weight  of  one  unit  in  length  of  the 
wall,  or,  for  a  trapezoid, 


(51)     | 


and  Vs  is  the  moment  of  that  weight  with  respect  to  the 
inner  edge  B  of  the  base.  By  considering  the  trapezoid 
ABCD  as  the  difference  between  the  rectangle  AaCc  and  the 
two  triangles  AaB  and  CcD,  this  is  found,  by  the  help  of  the 
principles  of  statics,  to  be 


Vs  =  \vh((t  +  0&  +  P  —  (2a  +  b)k  cot  0),      .     (52) 

and,  dividing  by  F,  the  distance  s  can  be  determined  if  it 
should  be  required.  To  investigate  a  wall,  formulas  (52)  and 
(51)  are  first  used,  then  (50),  and  lastly  (45). 


ART.  20.]    COMPRESSIVE   STRESSES  OF   THE  MASONRY.  63 

As  an  example,  let  the  following  data  be  taken :  A  sand- 
stone wall  retaining  a  level  bank  of  earth  ;  0—34  degrees, 
w  =  100  pounds  per  cubic  foot,  h  —  18  feet,  a  =  2  feet,  b  =  5 
feet,  6  •=.  80  degrees,  v  =  140  pounds  per  cubic  foot.  The 
value  of  P,  from  Article  8,  is  3570  pounds,  z  being  zero.  The 
weight  V  is  found  by  (51)  to  be  8820  pounds.  Vs  is  found  by 
(52)  to  be  4405  pounds-feet.  These  inserted  in  (50)  give 
t  =  1.75  feet,  which,  being  greater  than  one-third  the  base, 
shows  proper  stability;  and  lastly,  from  (48),  the  factor  of 
security  is  n  =  3.3. 

It  will  be  interesting  to  test  the  same  wall  by  the  pressure 
theory  of  Article  9,  where,  z  being  34  degrees,  P  is  2590 
pounds.  All  other  data  being  the  same  as  before,  there  is 
found  from  (50)  the  value  /  =  3.23  feet,  which  is  more  than 
one-half  the  base,  so  that  T  in  Figure  17  lies  between  M  and 
13,  and  the  tendency  to  rotation  about  B  is  greater  than  that 
about  C. 

Problem  19.  Compute  the  factor  of  security  against  rota 
tion  for  the  data  given  in  Problem  18. 


ARTICLE  20.    COMPRESSIVE  STRESSES  ON  THE  MASONRY. 

As  a  general  rule,  the  working  compressive  stress  upon  the 
base  of  masonry  walls  should  not  exceed  150  pounds  per 
square  inch  in  first-class  work.  A  tower  150  feet  in  height 
will  produce  this  pressure  on  its  base  if  the  masonry  weighs 
144  pounds  per  cubic  foot. 


64  INVESTIGATION  OF  RETAINING    WALLS.  [CHAP.  III. 

The  total  normal  pressure  N  upon  the  horizontal  base  of  a 
retaining  wall  will  be  given  by  (44),  making  a  =  o°,  or 

N=  r-Pcos(0  +  *X (53) 

in  which  P  is  the  earth-pressure  acting  at  an  angle  z  with  the 
normal  to  the  back  of  the  wall,  V  the  weight  of  the  wall,  and  0 
the  angle  which  its  back  makes  with  the  horizontal.  If  P  is 
computed  by  Article  8,  the  angle  z  is  zero ;  if  by  Article  9,  its, 
value  is  0,  the  angle  of  repose  of  the  earth.  For  any  ordinary 
case  cos  (0  +  z)  is  a  small  fraction,  and  in  most  cases  it  is 
a  sufficient  practical  approximation  to  regard  N  as  equal  to  V. 

The  compressive  stresses  upon  the  base  BC  (Figure  17)  will 
be  regarded  as  caused  by  the  vertical  pressure  N  alone.  N  is 
\  evidently  the  vertical  component  of  the  resultant  R.  The 
horizontal  component  of  R  produces  shearing  stresses  along 
the  base  which  are  supposed  not  to  increase  the  compressive 
stresses.  The  distribution  of  the  compression  over  the  base 
will  then  be  similar  to  that  shown  in  the  diagrams  of  Figure 
15,  and  will  depend  upon  the  position  of  the  point  in  which  R 
cuts  the  base. 

If  the  resultant  cuts  the  base  at  its  middle  point  (as  in  the 
first  diagram  of  Figure  15),  the  compression  due  to  N  is  uni- 
form over  the  area  b  X  I  square  feet,  and 

S=J^ (54) 

s**V 

is  the  compressive  stress  in  pounds  per  square  foot. 


ART.  20.]    COMPRESSIVE   STRESSES  OF   THE  MASONRY.          65 

If  the  resultant  is  applied  at  the  limit  of  the  middle  third 
(as  in  the  third  diagram  of  Figure  15),  the  unit-stress  at  the 
edge  B  is  zero,  that  at  the  middle  is  the  average  value  given 
by  (54),  and  the  greatest  stress_at  the  tQ£_6js  double  this  aver- 
age value,  or 

S=2y  ........    (55) 

If__the  resultant  is  applied  without  the  middle  third  at  a 
distance  /  from  the  edge  C  (as  in  the  last  diagram  of  Figure 
15),  the  compression  is  distributed  only  over  the  distance  3;, 
so  that 


(s6) 


gives  the  stress  in  jpmmdsjpei^square  foot. 


The  case  where  R  cuts  the  base  within  the  middle  third 
at  a  distance  /  from  C  (as  in  the  second  diagram  of  Figure  15) 
remains  to  be  considered.  Let  5  be  the  greatest  unit-stress  at 
(7,  and  5",  be  the  least  unit-stress  at  B.  Then  the  unit-stress  at 
the  middle  of  the  base  is  equal  to  the  average  unit-stress,  or 


and  as  N  is  applied  opposite  the  centre  of  gravity  of  the 
stress-trapezoid,  the  value  of  t  is 


_ 

~s+s,  v 


fa**. 


•: 


66  INVESTIGATION  OF  RETAINING    WALLS.   [CHAP.  III. 

Now  eliminating  ^  from  these  two  equations,  there  result 

v 


.  .   „ 

which  is  the   greatest  unit-stress,  namely,  that  at  the  toe  C.^h 
As  N  is  in  pounds  for  one  foot  in  length  of  wall,  and  b  is  in 
feet,  these  formulas  give  compressive  stresses  in  pounds  per 
square  foot,  and   dividing   by  144  the  values  in  pounds   per 
square  inch  are  found. 


^  example,  take  the  wall  of  the  last  article,  where  h  —  18 
feet,  a  =  2  feet,  b  —  5  feet,  6  =  80  degrees,  z  =  o°,  P=  3570 
pounds,  F=882O  pounds,  and  t  —  1.75  feet.  In  (44)  the 
value  of  a  is  o°,  and  N  is  found  to  be  8665  pounds.  Then 
from  (57)  the  greatest  compression  is  22.9  pounds  per  square 
inch,  which  is  a  low  value  even  for  inferior  work. 

For  ordinary  walls  a  sufficiently  exact  computation  of  the 
unit-stress  5  may  be  made  by  taking  V  for  the  value  of  N. 
Thus  for  the  above  case  V=  8820,  and  from  (57)  5=  23.3 
pounds  per  square  inch.  When  z  =  o°  and  a  =  o°,  formula 
(44)  gives  N  =  V  —  P  cos  0,  which  differs  but  little  from 
N  =•  Fwhen  0  is  near  90°. 

If  there  be  no  pressure  behind  a  wall,  the  point  T  coincides 
with  5  (Figure  17).  Then  the  normal  pressure  V  produces  the 
greatest  unit-stress  at  B,  whose  value  is  given  by  one  of  the 
formulas, 

'     '    (58) 


ART.  20.]    COMPRESSIVE   STRESSES  OF   THE  MASONRY.  6? 

according  as  the  distance  s  is  less  or  greater  than  one-third 
of  A 

According  to  the  theory  here  presented,  the  vertical  com- 
ponent of  R  alone  produces  compression  on  the  base  of  a 
retaining  wall,  while  the  horizontal  component  is  exerted  in 
producing  a  shearing  stress.  This  theory  has  defects  ;  but 
upon  it  has  been  based  the  design  of  structures  more  impor- 
tant than  retaining  walls. 

Problem  20.  Let  the  back  of  the  wall  be  inclined  forward 
at  a  batter  of  2  inches  per  foot,  and  let  the  normal  pressure  of 
the  earth  be  P=  22,760  pounds.  Let  its  height  be  36  feet, 
the  top  thickness  6  feet,  the  base  thickness  18  feet,  and  the 
weight  per  cubic  foot  150  pounds.  Compute  the  greatest 
compressive  stress  on  the  base. 


68  DESIGN  OF  RETAINING    WALLS.  [CHAP.  IV. 


CHAPTER    IV. 
DESIGN  OF   RETAINING  WALLS. 

ARTICLE  21.    DATA  AND  GENERAL  CONSIDERATIONS. 

When  a  retaining  wall  is  to  be  designed  for  a  particular 
location  the  character  of  the  earth  to  be  supported  is  known 
and  also  the  height  of  the  wall.  The  data  then  are  :  w  the 
weight  per  cubic  foot  of  the  earth,  <p  its  angle  of  repose,  6  the 
angle  of  inclination  of  its  surface,  and  h  the  height  of  the 
proposed  wall. 

The  thickness  of  the  top  of  the  wall,  a,  is  first  assumed. 
In  doing  this  practical  considerations  will  generally  govern 
rather  than  theoretical  ones.  Theory  indicates,  as  will  be 
seen  in  Article  24,  that  the  thinner  the  top  of  the  wall  the 
less  is  the  quantity  of  material  required  ;  but  theory  supposes 
the  earth  to  be  homogeneous  and  takes  no  cognizance  of 
the  action  of  frost.  Experience,  however,  teaches  that  the 
freezing  earth  near  the  top  of  the  wall  exerts  a  marked  lateral 
pressure  which  can  only  be  counteracted  by  a  substantial 
thickness.  To  possess  proper  stability  against  the  action  of 


ART.  21.]      DATA   AND   GENERAL   CON  SID  ERA  TIONS.  69 

frost  and  the  weather  a  wall  should  not  have  a  top  thick- 
ness less  than  two  feet.  Usually  when  the  height  of  a  wall 
varies,  as  in  a  railroad  cut,  the  top  has  the  same  thickness 
throughout.  If  a  wall  be  only  a  foot  high,  its  thickness 
should  not  be  less  than  two  feet,  else  in  a  few  years  the  frost 
will  push  it  over.  Even  in  latitudes  where  frost  is  rare  this 
rule  is  a  good  one  to  follow. 

The  engineer  will  next  decide  upon  the  batter  of  the  back 
of  the  wall,  or  upon  the  value  of  6,  the  angle  between  the  back 
and  the  horizontal.  In  doing  this  he  must  have  regard  to  the 
batter  which  the  front  of  the  wall  will  have,  and  to  the  theory 
of  economy  of  material  set  forth  in  the  following  articles.  In 
construction  the  back  of  the  wall  will  be  left  rough  or  built  in 
a  series  of  steps,  so  that  6  need  be  taken  only  to  the  nearest 
degree  of  the  average  inclination. 

The  pressure  of  the  earth  can  now  be  computed  by  the 
proper  formula  of  Chapter  II.  The  theory  of  Article  8  which 
supposes  its  direction  to  be  normal  to  the  back  of  the  wall  is, 
in  general,  to  be  preferred,  because  in  practice  the  earth  is 
tamped  against  the  wall  so  that  there  can  be  little  tendency 
to  slide  along  it.  Article  8  demands  a  heavier  wall  than 
Article  9,  and  is  thus  on  the  side  of  safety.  In  our  opinion 
Article  8  gives  the  pressure  of  earth  against  a  wall  which 
stands  firmly  with  a  high  degree  of  stability,  and  Article  9 
gives  the  pressure  of  the  earth  when  motion  or  failure  is  about 
to  begin.  As  walls  are  designed  to  stand  and  not  to  fail,  the 
engineer  should  be  careful  in  erring  on  the  side  of  safety. 


70  DESIGN  OF  RETAINING    WALLS.  [CHAP.  IV. 

Therefore  in  this  chapter  the  lateral  pressure  P  will  usually  be 
taken  normal  to  the  back  of  the  wall,  so  that  in  all  previous 
formulas  the  angle  z  is  zero. 

The  next  procedure  is  to  determine  the  thickness  of  the 
base  so  that  the  wall  may  have  proper  security  against  rota- 
tion ;  how  this  is  done  the  following  Article  will  show.  If  the 
front  of  the  wall  thus  designed  does  not  have  the  desired 
batter,  a  change  can  be  made  in  the  value  of  6  and  the  work 
be  repeated.  Then  it  will  be  well  to  test  the  work  by  deter- 
mining graphically  (Article  18)  the  factor  of  security  against 
rotation.  Lastly,  the  question  of  sliding  must  be  considered 
and  proper  security  against  it  be  provided  (Article  23).  The 
practical  points  regarding  the  coping,  the  frost  batter  near  the 
top  of  the  back  of  the  wall,  the  weep  holes,  the  foundation, 
the  drainage  ditches,  the  quality  of  the  masonry,  and  the 
details  of  construction  will,  of  course,  receive  full  attention 
and  be  fully  set  forth  in  the  drawings  and  specifications. 

Problem  21.  If  ft  be  the  angle  which  the  front  of  the 
wall  makes  with  the  horizontal,  prove  that  b  —  a  equals 
h  (cot  ft  —  cot  0).  Find  the  batter  of  both  back  and  front 
in  inches  per  foot  when  b  =  5  feet,  a  —  2  feet,  h  =  18  feet,  and 
6  =  80  degrees. 


ART.  22.]  COMPUTATION  OF   THICKNESS.  71 


ARTICLE  22.    COMPUTATION  OF  THICKNESS,  y 

,• 

The  discussion  in  Article  19  furnishes  the  following  funda- 
mental equation  for  the  stability  of  any  wall  against  rotation  : 


To  apply  this  to  the  determination  of  the  thickness  of  the 
base  of  a  trapezoidal  wall  the  values  of  /,  V  and  Vs  are 
inserted  from  (49),  (51)  and  (52)  and  t  is  made  \b,  thus  giving 
a  factor  of  security  of  3.0  against  rotation  (Article  17).  The 
value  of  the  angle  z  is  taken  as  zero  because  the  earth  pressure 
is  computed  from  Article  8  under  that  supposition.  Then 
results 


b  cos  B  +  =  &h(P  +  «*-«"  +  bh  cot0+  2«*cot0),  (59) 


and  the  solution  of  tljis  equation  with  respect  to  b  gives 

b  =  -  A  +  V£  +  A\ (60) 

in  which  A  and  B  have  the  values 

_  _  2/>,  cos  9 

vh      ' 


B  =  --  + a*  -  2ah  cot  B, 
v  sin  u 


7  2  DESIGN  OF  RETAINING    WALLS.  [CHAP.  IV. 

from   which    the   base    thickness  can    be    computed    for   any 
given  data. 

When  the  value  of  0  is  90  degrees  this  takes  the  simple 
form 


(61) 


which  is  the   formula  for  the  b***e  thickness  of   a  wall  with 
vertical  back. 

In  these  formulas  Pl  is  the  earth  pressure  computed  by 
Article  8,  h  the  vertical  height  of  the  wall,  a  the  thickness  of 
its  top,  6  the  angle  at  which  the  back  is  inclined  to  the  hori- 
zontal, v  the  weight  of  a  cubic  foot  of  masonry,  and  b  is  the 
thickness  of  the  base  which  gives  the  wall  a  factor  of  security 
of  3.0  against  rotation,  the  resultant  R  then  cutting  the  base 
at  the  limit  of  the  middle  third.  For  all  joints  above  the  base 
the  factor  of  security  will  then  be  greater  than  3.0. 

For  example,  let  a  wall  with  vertical  back  be  20  feet  high, 
sustaining  a  level  bank  of  sand  which  weighs  100  pounds  per 
cubic  foot  and  has  a  natural  slope  of  34  degrees.  Let  the 
masonry  be  165  pounds  per  cubic  foot  and  the  top  of  the  wall 
be  2  feet  in  thickness.  It  is  required  to  find  the  thickness  of 
the  base  BC  (Figure  16).  From  formula  (26)  the  pressure  of 
the  earth  is  5650  pounds.  Then  from  (61) 


ART.  22.]  COMPUTATION  OF   THICKNESS.  73 

which  gives  a  cross-section  whose  area  is  K2  4~  7-57)x£  or 
95.7  square  feet. 

As  a  second  example  take  the  same  wall  except  that  the 
back  is  inclined  backward  so  that  9  is  80  degrees.  Here  the 
value  of  Pl  is  found  from  (25)  to  be  4410  pounds.  Then 

whence  from  (60)  A  =  229,  B  =  44.2,  and  b  =  4.45  feet, 
which  gives  a  cross-section  whose  area  is  64.5  square  feet. 
The  advantage  of  inclining  the  wall  backward  is  here  plainly 
indicated,  the  vertical  wall  requiring  nearly  50  per  cent  more 
material  than  the  inclined  one. 

If  the  wall  be  of  uniform  thickness  throughout,  a  equals  b, 
and  the  solution  of  (59)  gives 


(62) 


in  which  C  has  the  value 

r_  3/z  cot  6      2/>,  cosfl 
2  ~~vh      ' 

If  in  this  6  be  90  degrees,  it  becomes 


which  is  the  proper  thickness  for  a  vertical  rectangular  wall. 
As  an  illustration  take  the  same  bank  of  sand  as  in  the  last  ex- 
ample ;  then  for  8  —  80°,  C  '  =  4.81  and  the  required  thickness 
is  £='4.0  feet.  If,  however,  6  =  90  degrees,  there  is  found 


74 


DESIGN  OF  RETAINING    WALLS. 


[CHAP. 


b  =  8.3   feet.      Here  again  the  great  advantage  of  inclining 
the  wall  is  seen. 

Sometimes  it  may  be  desirable  to  assume  the  inclination  fi 
of  the  front  of  the  wall,  and  then  to  compute  both  b  and  a. 
For  this  case  Figure  17  gives 

a  —  b—  //(cot  /?  —  cot  0),        ....     (63) 
and  inserting  this  in  (59)  and  solving  for  b  there  is  found 

.....     (64) 


in  which  D  and  E  are  determined  from 


D  =  /Kcot  6  +  j-  cot  ft)  -       - 


B 


vh 


•p 


z>sm 


For  example,  take  the  same  bank  of  sand  as  before  and  let  the 
back  be  vertical,  or  6  =  90°,  and  h  =  2O  feet.  Then  Pl  =  5650 
pounds  per  linear  foot  of  wall.  Now  let  the  front  of  the  wall 
have  the  batter  of  i-J-  inches  per  foot,  or  /3  =  82°  52',  and  cot  ft 
=  0.1250  (Article  13).  Then  Z>  =  1.25  and  £  =  68.5  and 
from  (64)  the  base  thickness  is  b  =  7.12  feet ;  lastly  from  (63) 
the  top  thickness  is  a  —  5.87  feet. 

The  formulas  above  given  can  only  be  used  when  the  earth 

^pressure  Pl  has  a  direction  normal  to  the  back  of   the  wall. 

Those  who  believe  in  the  theory  of  earth  pressure  set  forth  in 

Article  9  are  referred  to  the  latter  part  of   Article  24  for  a 

formula  by  which  they  should  compute  the  thickness. 


ART.  23.]  SECURITY  AGAINST  SLIDING.  75 

Problem  22.  A  wall  weighing  140  pounds  per  cubic  foot 
has  a  vertical  back,  is  18  feet  high,  and  the  horizontal  earth 
pressure  on  it  is  4580  pounds.  Compute  the  thickness  of  the 
base  when  the  cross-section  is  rectangular.  Compute  the 
thickness  of  the  base  when  the  cross-section  is  triangular. 
Compare  the  two  sections  with  respect  to  amount  of  material. 


ARTICLE  23.     SECURITY  AGAINST  SLIDING. 
(p-n//C^~a  -jLl&iftr^l 

The  base  thickness  b  computed  in  the  last  Article  provides 
proper  security  against  the  rotation  of  the  wall  under  the  lat- 
eral pressure  of  the  earth.  The  cross-section  thus  determined 
should  now  be  investigated  and  full  security  against  sliding  be 
provided.  This  can  be  done  in  three  ways. 

First :  the  masonry  "may  be  laid  with  random  courses  so 
that  no  through  joints  will  exist.  If  the  stones  are  of  suffi- 
cient size  this  checks  very  effectually  all  liability  to  sliding. 

Second  :  all  through  joints  may  be  inclined  backward  at  an 
angle  a  (Fig.  13)  so  that  the  resultant  R  shall  be  as  nearly  nor- 
mal to  them  as  possible.  This  will  occur  when  Fin  formula 
(44)  is  zero,  or  when 

/>,  sin  (#  +  «)=  Fsin  a, 
and  this  reduces  to 

cot  a  =        V       -  cot  0,       ....     (65) 
P.  sm  6 


76  DESIGN  OF  RETAINING    WALLS.  [CHAP.  IV. 

from  which  a  can  be  computed  for  any  joint,  V  being  the 
weight  of  the  wall  above  that  joint,  Pl  the  earth  pressure  above 
it,  and  0  the  inclination  to  the  horizontal  of  the  back  of  the 
wall.  As  b  is  computed  for  a  horizontal  base,  the  value  of  Fis 
a  little  less  than  \vh  (a  -f-  b).  For  example,  take  the  wall  de- 
signed above  where  B  =  90  degrees,  //  =  20  feet,  P^  =  5650 
pounds,  ^=165  pounds  per  cubic  foot,  a  —  5.9  feet,  and 
b=  7.1  feet.  Then  Fis  a  little  less  than  21  450  pounds,  say 
21  ooo  pounds,  and  cot  a  =  2.7,  which  gives  a  =.  20  degrees 
nearly.  This  backward  inclination  should  be  made  less  for 
joints  above  the  base,  becoming  nearly  zero  for  those  near  the 
top  of  the  wall. 

Third  :  for  cases  where  a  through  horizontal  joint  cannot 
be  avoided,  as  when  a  wall  is  built  on  a  platform,  the  thickness 
of  the  base  which  will  give  a  required  factor  of  security  against 
sliding  can  be  computed  from  (43).  To  do  this  make  both  z 
and  a  equal  to  zero  in  (44),  and  substitute  the  values  of  F  and 
TV  from  (43),  giving 

*/>1sin0=./(F-/>cos0). 

Now  in  this  let  the  value  of  V  be  inserted,  namely, 

V=  kvh(a+  b\ 
and  the  equation  be  solved  for  b,  thus : 

*2«±f™*l,     ...    (66) 


ART.  24.]  ECONOMIC  PROPORTIONS.  77 

in  which  a  is  the  top  thickness,  h  the  vertical  height,  0  the  in- 
clination of  the  back  of  the  wall  to  the  horizontal,  P^  the  nor- 
mal pressure  of  the  earth,  v  the  weight  of  the  masonry  per 
cubic  unit, /the  coefficient  of  friction  of  the  masonry  on  the 
through  horizontal  joint,  and  b  the  base  thickness  for  a  factor 
of  security  of  n  against  sliding,  It  would  be  desirable  that  n 
should  be  about  3.0,  but  to  secure  this  the  wall  must  be  thicker 
than  is  required  for  rotation.  Accordingly,  this  method  of 
obtaining  security  against  sliding  should  be  used  only  when  all 
other  methods  are  impracticable.  Thus  in  the  last  article  a 
vertical  rectangular  wall  is  determined  to  be  8.34  feet  thick 
when  k=.  20  feet  and  Pl  =  5650  pounds,  v  =  165  pounds  per 
cubic  foot;  now,  if  n  =3.0  and  /=  0.5,  formula  (66)  gives 
a  =  b  —  10.3  feet. 

Problem  23.  Compute  the  proper  inclination  of  the  joints 
in  the  rectangular  wall  of  Problem  22  at  distances  of  6,  12  and 
1 8  feet  from  the  top. 


ARTICLE  24.    ECONOMIC  PROPORTIONS. 

By  the  help  of  the  formulas  of  Article  22  the  thicknesses 
of  several  trapezoidal  walls  will  now  be  computed  in  order  to 
compare  the  quantities  of  masonry  required,  and  thus  obtain 
knowledge  regarding  the  most  economical  forms  of  cross-sec- 
tion. All  the  walls  will  be  18  feet  in  vertical  height,  and 
sustain  a  level  bank  of  earth  whose  angle  of  repose  is  34  de- 
grees and  which  weighs  100  pounds  per  cubic  foot.  The 


7%  DESIGN  OF  RETAINING    WALLS.  [CHAP.  IV. 

weight  of  the  masonry  will  be  taken  as  150  pounds  per  cubic 
foot. 

Case  I. — Let  the  back  of  the  wall  be  inclined  forward  at  a 
batter  of  two  inches  per  foot,  or  6=  99°  28'  (Fig.  18).  From 
formula  (25)  the  normal  earth  pressure  Pl  is  found  to  be  5690 
pounds.  Then  assuming  the  top  thickness  a  at  o.o,  i.o,  2.0 
feet,  etc.,  the  proper  base  thickness  for  each  is  computed  from 
formula  (60)  and  given  in  the  table  below. 

Case  II. — Let  the  back  of  the  wall  be  vertical,  as  in  Fig.  19, 
or  6  =  90°.  From  formula  (26)  the  earth  pressure  Pl  is  found 


FIG.  18. 

to  be  4580  pounds.  Then  assuming  thicknesses  of  the  top,  the 
corresponding  base  thicknesses  are  computed  and  inserted  in 
the  following  table. 

In  this  table  the  column  headed  "cubic  yards"  gives  the 
quantity  of  masonry  in  one  linear  foot  of  the  wall,  and  it  is 
seen  that  in  each  case  this  is  least  for  the  wall  with  the  thin- 
nest top.  It  is  also  seen  that  the  vertical  walls  require  less 
masonry  than  the  corresponding  ones  with  forward  batters. 
The  columns  headed  "  per  cent  "  show  these  facts  more  clearly, 
the  standard  of  comparison  being  the  vertical  rectangular  wall 
which  is  taken  as  100. 


ART.  24.] 


ECONOMIC  PROPORTIONS. 


79 


Assumed 

T*/\f\ 

Case  I.    9  =  99°  28'. 

Case  II.    9  =  90°. 

1  Op 

Thickness. 
a. 

Base 
Thickness. 
*. 

Cubic 
Yards. 

Per  cent. 

Base 
Thickness. 
b 

Cubic 
Yards. 

Per  cent. 

Feet. 

Feet. 

Feet. 

0.0 

9.6 

3.20 

62 

7.8 

2.60 

50 

I.O 

9-5 

3-50 

67 

7-3 

2.77 

53 

2.0 

9-4 

3.80 

73 

7-1 

3-03 

58 

3-0 

9-5 

4.17 

80 

7-1 

3-37 

65 

4.0 

9.6 

4-57 

88 

7-1 

3-70 

71 

5-o 

9.9 

4-97 

96 

7-1 

4-03 

77i 

6.0 

10.2 

5-40 

104 

7.2 

4.40 

85 

7.0 

10.5 

5.83 

112 

7-5 

4.83 

93 

7.8 

7-8 

5.20 

IOO 

7-9 

lO.g 

6.27 

120 

Case  III. — Let  the  back  of  the  wall  be  inclined  backward 
at  a  batter  of  ij  inches  per  foot,  or  0  =  82°  53'  (Fig.  20). 
Here  the  earth  pressure  Pl  is  found  to  be  3850  pounds.  Then 


FIG.  20. 


FIG.  ax. 


assuming  values  of  the  top  thickness  a,  the  corresponding 
values  of  the  base  thickness  b  are  computed  from  (60)  and 
given  below  in  the  tabulation. 

Case  IV. — Lastly,  let  the  back  of  the  wall  be  inclined  still 
more  backward,  the  batter  being  3  inches  per  foot,  or  B  =?$°  58', 
as  in  Fig.  21.  Then  the  earth  pressure  is  found  to  be  3200 


8o 


DESIGN  OF  RETAINING   WALLS. 


[CHAP.  IV. 


pounds,  and  as  before  values  of  b  are  computed  for  assumed 
values  of  a. 

The  following  table  gives  the  results  of  these  computations 
for  Cases  III  and  IV,  the  columns  "cubic  yards"  and  "per 
cent "  having  the  same  signification  as  before.  It  is  seen  that 
the  general  laws  of  economy  of  material  are  the  same,  namely,, 


Assumed 

Case  III.    0  =  82°  53'. 

Case  IV.    0  =  75°  58'. 

Top 
Thickness, 
a. 

Base 
Thickness. 
b. 

Cubic 
Yards. 

Per  cent. 

Base 
Thickness. 
b 

Cubic 

Yards. 

Per  cent. 

Feet. 

Feet. 

Feet. 

0.0 

6.6 

2.20 

42 

5-1 

1.70 

33 

1.0 

5-9 

2-30 

44 

4.2 

1-73 

33 

2.0 

5-4 

2.47 

47* 

3-4 

1.  80 

35 

2.9 

2.9 

1-93 

37 

3-0 

5-1 

2.70 

52 

4-0 

4-9 

2.97 

57 

4-9 

4.9 

3-30 

63 

the  thinner  the  top  and  the  greater  the  backward  batter  of 
the  wall  the  less  is  the  quantity  of  masonry.  The  considera- 
tion of  these  principles  in  connection  with  the  local  circum- 
stances of  an  actual  case  will  hence  tend  toward  economy  of 
construction.  Chief  among  these  local  circumstances  is  the 
price  of  land,  and  where  this  is  very  high  a  wall  with  a  verti- 
cal front  and  a  forward  batter  of  back  is  often  used,  al- 
though this  requires  more  masonry  than  any  other  form,  for 
the  saving  in  cost  of  the  land  may  more  than  balance  the 
extra  expense  for  masonry.  In  all  cases  of  design  the  first 
consideration  is  stability,  and  the  second  economy — not  econ- 


ART.  24.]  ECONOMIC  PROPORTIONS.  8 1 

omy  in  the  cost  of  material,  but  in  the  total  expenditure  of 
money.  

Those  who  believe  in  the  theory  of  earth  pressure  set  forth 
in  Article  9  may  ask  if  its  use  would  lead  to  the  same  conclu- 
sions regarding  economic  proportions.  To  decide  this  it  *is 
necessary  to  deduce  a  formula  for  the  thickness  of  a  trape- 
zoidal wall  under  such  pressure,  and  then  to  make  the  same 
computations  for  the  four  cases  with  the  same  data. 

The  fundamental  formula  (47)  is  good  for  all  cases.  In 
this  let  the  values  of/,  F,  and  Vs  be  inserted  from  (49),  (51) 
and  (52),  making  z  =  0  and  t  =  \b.  Then  results 


Pa[2&  cos  (9+  0)  +  — ^-- ]  =  \vh(P  +  ab-ai  +  bh  cot  0  -f  2ah  cot  6), 
\  sin  6   / 

arid  solving  this  with  respect  to  b  there  is  found 


b  =  -A+  VB  +  A*, (67) 

in  which  the  values  of  A  and  B  are 


„        2P~  COS  0     .  , 

B  = — H-- — jr-  +  #2  —  2^  cot  0, 
z/  sin  0 

and  from  this  the  base  thickness  b  can  be  computed  for  any 
values  of  the  given  data,  namely,  the  angle  of  repose  of  the 
earth  0,  its  inclined  pressure  Pa  as  found  by  Article  9,  the 


82 


DESIGN   OF  RETAINING    WALLS. 


[CHAP.  IV. 


angle  of  inclination  of  the  back  of  the  wall  6,  the  top  thick- 
ness a,  the  vertical  height  7z,  and  its  weight  per  cubic  unit  v. 

Using  the  same  data,  the  inclined  pressure  P^  has  been 
computed  for  each  case,  and  the  base  thicknesses  found  from 
formula  (67)  for  the  same  assumed  top  thicknesses.  The 
cubic  yards  in  one  linear  foot  of  wall  are  next  obtained,  and 
an  inspection  of  these  shows  that  the  same  general  laws  hold 
as  before,  namely,  the  thinner  the  wall  and  the  less  the  angle 
6  the  less  is  the  quantity  of  masonry  required. 

The  subjoined  table  gives  the  quantities  of  masonry  for 
Case  I,  Case  II,  and  Case  IV,  and  by  comparing  them  with 


Assumed 
Top 
Thickness. 
a. 

Case  I.     0  =  99°  28'. 

Case  II.     0  =  90°. 

Case  IV.     0  =  75°  58'. 

Cubic 
Yards. 

Per  cent 
Difference. 

Cubic 
Yards. 

Per  cent 
Difference. 

Cubic 
Yards. 

Per  cent. 
Difference. 

Feet, 
o.o 

2.43 

24 

1-77 

32 

1.38 

19 

1.0 

2.77 

21 

2.00 

28 

1.40 

J9 

2.0 

3.10 

19 

2-30 

24 

1-45 

19 

3-0 

3-50 

16 

2.63 

22 

4.0 

3-97 

13 

3-00 

19 

5-o 

4-50 

9 

3-40 

15 

6.0 

5-07 

6 

those  previously  deduced  it  is  seen  that  they  are  all  less,  the 
difference  being  greatest  for  the  triangular  walls  and  least  for 
those  of  uniform  thickness.  The  column  "per  cent  difference" 
shows  in  each  case  the  percentage  of  material  which  the  walls 
designed  under  inclined  pressure  are  less  than  the  correspond- 


ART.  25.]  THE  LINE    OF  RESISTANCE.  83* 

ing  ones  designed  under  normal  pressure.  As  in  practice 
walls  are  not  built  with  a  top  thickness  less  than  two  feet,  it 
may  be  said  as  a  rough  rule  that  the  hypothesis  of  inclined 
earth  pressure  (Article  9)  gives  a  wall  from  10  to  20  per  cent 
less  in  size  than  that  of  normal  earth  pressure  (Article  8). 

Problem  24.  Deduce  a  formula  for  the  thickness  of  a  wall 
under  inclined  earth  pressure  when  a  =  b.  Compute  the 
thickness  and  quantity  of  material  of  such  a  wall  for  Case  I, 
for  Case  II,  and  for  Case  IV. 


ARTICLE  25.    THE  LINE  OF  RESISTANCE. 

Let  a  be  the  top  thickness  and  b  the  base  thickness  of  a 
trapezoidal  wall  whose  height  is  h.  Then  the  thickness  b1  at  a 
vertical  distance  y  below  the  top  is 


a),     .....       (68) 
and  this  is  represented  by  B'C'  in  Figure  22.     Let  P  be  the 


*— a- 


FIG 


pressure  of  the  earth,  and    V  the  weight  of  the  wall  above 
B'C.     Let  T'  be  the  point  where  the  resultant  of  P  and  V 


DESIGN  OF  RETAINING    WALLS. 


[CHAP. 


cuts  B 'C ';  as  y  varies  T'  describes  a  curve  called  the  line  of 
resistance.  When  y  is  zero  T'  coincides  with  the  middle  of  the 
top.  When  y  equals  h  the  point  T'  coincides  with  T  as  de- 
termined by  (50). 

The  line  of  resistance  is  the  locus  of  the  point  of  intersec- 
tion of  the  resultant  of  the  forces  above  any  horizontal  joint 
with  the  plane  of  that  joint.  This  is  a  general  definition 
applicable  to  triangular  and  curved  sections  as  well  as  to  trape- 
zodial  ones. 

For  a  rectangular  vertical  wall  under  normal  earth  pressure 
the  line  of  resistance  is  the  common  parabola.  To  prove  this 
let  the  origin  of  coordinates  be  taken  at  the  corner  A  in  Figure 
-24,  and  let  AB'  =y  and  B'T'  =  x.  Now  P'  =  cy\  in  which  c 


FIG.  23. 

is  a  function  of  w,  <f>  and  d  (Article  8),  and  its  lever-arm  with 
respect  to  T'  is  \y.  The  value  of  Fis  vby,  and  its  arm  with 
respect  to  T'  is  x  —  \b.  Then  the  equation  of  moments  is 


or 


ART.  25.]  THE  LINE   OF  RESISTANCE.  8$ 

which  represents  a  parabola  with  its  vertex  at  the  middle  of 
the  top  of  the  wall. 

For  a  triangular  wall  with  a  vertical  back  the  line  of  resist- 
ance is  a  straight  line  drawn  from  the  top  to  the  point  where 
the  resultant  cuts  the  base.  The  proof  of  this  is  purposely 
omitted  in  order  that  it  may  be  worked  out  by  the  student. 

For  a  trapezoidal  section  the  position  of  the  line  of  resist- 
ance can  be  computed  from  (50),  (51)  and  (52),  making  z  =  o 
for  normal  earth  pressure,  putting  P=  rj/a,  h  =  y  and  b  =  bf  . 
For  example,  take  a  wall  for  which  0  =  34°,  d  =  o°,  w  =  100, 
h  —  1  8,  9  —  '80°,  v  —  140,  a  =  2  and  b  —  5  feet.  Here  from 
formula  (25)  P  is  found  to  be  11.027".  From  (68) 


and  this  inserted  in  (51)  and  (52)  gives  the  values  of  Kand  Vs 
in  terms  of/.     Then  substituting  all  in  (50)  there  is  found 

280  +  67.487-  2.393?' 


28o  +  9-757- 
From  this  equation  the  curve  is  now  easily  constructed,  thus-* 

y  =    o,  t  —  i.oo,  and  b'  =  2.00 

y=    6,  £=1.77,  and  £'  =  3.00 

y  ~  12,  7=1.88,  and  ^  =  4.00 

7—18,  *=i.8i,  and  £'  =  5.00 

and  it  is  seen  that  the  line,  while  lying  always  within  the  mid- 
dle third,  departs  most  widely  from  the  middle  at  the  base  of 
the  wall. 


86  DESIGN  OF  RETAINING    WALLS.  [CHAP.  IV. 

Whatever  be  the  form  of  cross-section  the  line  of  resist- 
ance can  always  be  located  by  first  determining  the  earth 
pressure  and  the  weight  of  the  wall  for  several  values  of  y  and 
then  for  each  making  a  graphical  construction  as  in  Figure  12. 
The  curve  joining  the  points  thus  found  on  the  several  hori- 
zontal joints  will  be  the  line  of  resistance,  and  to  insure  proper 
stability  against  rotation  it  should  lie  within  the  middle  third 
of  the  wall  (Article  17). 

Problem  25.  Locate  graphically  the  line  of  resistance  in 
one  of  the  walls  of  Case  II,  Article  24,  determining  points  at 
depths  of  6,  12  and  18  feet  below  the  top. 


ARTICLE  26.    DESIGN  OF  A  POLYGONAL  SECTION. 

Retaining  walls  with  curved  front  are  now  and  then  built. 
The  advantages  claimed  for  such  a  profile  are,  first,  finer 
architectural  effect,  and  second,  that  the  line  of  resistance 
may  be  made  to  run  nearly  parallel  to  the  central  line  of  the 
wall,  thus  making  it  a  form  of  uniform  strength  and  insuring 
economy  of  material. 

The  determination  of  the  equation  of  a  curved  profile  to 
fulfil  the  condition  that  the  line  of  resistance  shall  cut  every 
joint  at  the  same  fractional  part  of  its  length  from  the  edge  is 
of  very  great  mathematical  difficulty,  if  not  impossibility,  be- 
cause the  weight  of  the  wall  above  any  joint  and  its  lever-arm 
are  unknown  functions  of  the  coordinates  of  the  unknown 
curve.  By  considering  the  curve  to  be  made  up  of  a  number 


ART.  2C.]  DESIGN  OF  A    POLYGONAL   SECTION.  87 

of  straight  lines,  however,  it  is  easy  to  arrange  a  profile  to  sat- 
isfy the  imposed  conditions  which  will  not  practically  differ 
from  the  theoretical  curve.  The  method  of  doing  this  will 
now  be  illustrated  by  a  numerical  example. 

A  wall  30  feet  in  vertical  height  is  to  be  designed  to  sup- 
port a  level  bank  of  earth  whose  angle  of  natural  slope  is  34 
degrees  and  which  weighs  100  pounds  per  cubic  foot.  The 
back  of  the  wall  is  to  be  plane  and  to  have  an  inclination  of  80 
degrees.  The  top  of  the  wall  is  to  be  2  feet  thick,  and  the 
weight  of  the  masonry  is  to  be  165  pounds  per  cubic  foot.  It 


A p 


FIG.  24. 


is  required  to  design  the  wall  so  that  the  line  of  resistance 
shall  cut  the  base  B£*  at  its  middle  point,  and  also  cut  the 
lines  B£i  and  BC  at  their  middle  points,  B1C1  being  20  feet 
and  BC  10  feet  from  the  top.  This  insures  a  factor  of  infinity 
against  rotation  (Article  17)  which  is  a  greater  degree  of  sta- 
bility than  is  usually  required  in  practice,  but  the  method  em- 
ployed is  general,  and  the  example  will  serve  to  show  how  a 
wall  may  be  designed  to  satisfy  any  imposed  condition 


DESIGN  OF  RETAINING    WALLS.  [CHAP.  IV. 

First,  take  the  upper  part  ABCD  and  consider  it  as  a  sim- 
ple trapezoidal  wall,  upon  which  the  normal  earth  pressure  is 
found  by  (25)  to  be  1410  pounds.  In  the  general  formula  (47) 
the  values  of  /,  V  and  Vs  are  now  to  be  substituted  from  (49), 
(51)  and  (52),  making  z  =  o  and  putting  /  equal  to  \b.  This 
gives  an  equation  in  which  all  quantities  but  b  are  known,  and 
by  its  solution  there  is  found  the  value  b  =  4.47  feet.  This 
completely  determines  the  cross-section  ABCD  so  that  it  is 
easy  to  find  the  weight  V  =  5240  pounds,  and  from  (52)  its 
lever-arm  s  =  1.55  feet. 

Second,  take  the  trapezoid  BCC^B^  and  consider  it  as  acted 
upon  by  four  forces,  the  weight  of  the  upper  part  5240  pounds, 
its  own  weight  V,  the  normal  pressure  of  20  feet  of  earth 


which  is  5650  pounds  acting  at  6f  feet  vertically  above  Bl , 
and  the  reaction  R  of  the  wall  below  it  which  by  the  hypoth- 
esis passes  through  M,  the  middle  point  of  B£r  Let  s  be 
the  lever-arm  of  Fwith  respect  to  £>,  and  let  BlCl  be  denoted 
by  b.  With  respect  to  the  centre  M  the  lever-arm  of  V  is 


ART.  26.]  DESIGN  OF  A   POLYGONAL   SECTION.  89 


\b  —  s,  that  of  the  5240  pounds  is  \b  +  0.21,  and  that  of  the 
earth  pressure  is  6.77  -f-  0.087$.  Then  the  equation  of  mo- 
ments is 

5650(6.77  +  0.087$)  =  V(kb  -  s)  +  524o(i$  +  0.21). 


Inse/ting  in  this  the  values  of  Fand  Vs  in  terms  of  b,  and 
then  solving,  there  is  found  b  —  8.70  feet.  This  determines  the 
cross-section  so  that  its  weight  V  is  found  to  be  108.50  pounds, 
and  the  lever-arm  of  this  with  respect  to  Bl  to  be  2.62  feet. 


Lastly,  the  trapezoid  B^Cf^JB^  is  treated  in  a  similar  man- 
ner, as  acted  upon  by  five  forces,  the  weights  5240  and  10850 
pounds,  the  pressure  of  30  feet  of  earth  which  is  12  720  pounds 
applied  at  Bl  ,  its  own  unknown  weight  F,  and  the  reaction  R 
which  passes  through  the  middle  of  Bfv  The  lever-arms  of 
the  known  forces  with  respect  to  that  centre  being  found,  the 
equation  of  moments  is 

• 
12720(10.15  +  0.087^)  =  V(\b  —  s)  +  5240^  +  1.97)  -}-  10850(1^  —  0.86), 

in  which  b  is  the  base  JB9C99  and  s  is  the  lever-arm  of  Fwith  re- 
spect to  Bv  From  (51)  and  (52)  the  values  of  Fand  Vs  are  to 
be  expressed  in  terms  of  b  and  inserted  ;  then  by  solution 
there  is  found  b  —  13.6  feet. 

The  points  C,  Cl  and  C^  in  the  profile  of  the  cross-section 
are  now  known,  and  a  curve  may  be  drawn  through  them,  or 
the  front  may  be  built  with  straight  lines.  The  economy  of 
the  curved  profile  is  indicated  by  the  fact  that  the  cross-section 
as  determined  is  209  square  feet,  whereas  a  trapezoidal  section, 


90  DESIGN  OF  RETAINING    WALLS.  [CHAP.  IV. 

designed  under  the  same  conditions  has  a  base  thickness  of 
15.2  feet  and  a  cross-section  of  257  square  feet. 

Problem  26.  Design  a  curved  wall  for  the  same  data  as 
above,  but  under  the  condition  that  the  line  of  resistance  shall 
cut  each  of  the  bases  BC,  B^C^Bf^  at  one-third  its  length  from 
the  outer  edge. 


ARTICLE  27.     DESIGN  AND  CONSTRUCTION. 

When  a  retaining  wall  is  to  be  designed  its  vertical  height 
will  be  given.  The  inclination  of  its  back  and  the  thickness  of 
its  top  are  to  be  assumed,  in  accordance  with  the  principles  of 
Article  24,  so  as  to  result  in  the  least  total  expenditure  for 
land,  labor  and  material.  The  form  of  section  selected  will  be 
usually  trapezoidal. 

The  normal  earth  pressure  is  now  computed  by  the  proper 
formula  of  Article  8. 

The  thickness  at  the  base  is  then  computed  by  formula  (60),. 
and  thus  the  cross-section  of  a  trapezoidal  wall  is  determined. 
The  batter  of  the  front  of  the  wall  is  known  by  (63),  and  if 
this  proves  to  be  greater  or  less  than  is  thought  advisable  new 
proportions  are  assumed  and  another  cross-section  determined. 

By  the  help  of  formula  (65)  the  approximate  inclination  of 
a  few  of  the  joints  should  next  be  found  so  that  the  wall  may 
be  built  with  full  security  against  sliding.  It  is  not  always 


ART.  27.]  DESIGN  AND   CONSTRUCTION.  gi 

necessary  to  give  the  joints  the  full  inclination  thus  computed, 
however,  since  this  implies  a  factor  of  security  of  infinity. 

As  a  check  on  the  computations  it  is  well  to  make  a  graph- 
ical investigation  of  the  proposed  wall  and  determine  the  fac- 
tors of  security  at  the  base  against  rotation  (Article  18)  and 
also  against  sliding  (Article  15).  These  will,  in  general,  be 
less  for  the  base  than  for  any  joint  above  the  base. 

Lastly,  the  maximum  pressure  per  square  inch  at  the  edge 
of  the  base  joint  may  be  computed  (Article  20).  If  this  is  less 


than  the  allowable  working  strength  of  the  stone,  the  wall  is 
safe  against  crushing.  Only  for  very  high  walls  will  this  com- 
putation be  necessary. 

The  computed  thickness  b  is  the  horizontal  thickness  of  the 
wall  at  the  top  of  the  foundation,  as  BC  in  Figure  22.  This 
foundation  should  be  built  with  care,  not  only  to  bear  the 
weight  of  the  wall  and  prevent  it  from  sliding,  but  also  to  pro- 


92  DESIGN  OF  RETAINING    WALLS.  [CHAP.  IV. 

tect  it  from  the  action  of  the  rain  and  frost.  Provision  should 
be  made  for  the  drainage  of  the  bank  by  longitudinal  ditches 
and  by  weep-holes  through  the  wall, so  that  water  may  not  col- 
lect and  increase  the  pressure. 

It  is  good  practice  to  batter  the  back  of  the  wall  slightly 
forward  for  about  two  feet  near  the  top,  in  order  that  the  frost 
may  lift  the  earth  upward  without  exerting  lateral  pressure 
against  the  wall. 

Whether  the  wall  be  built  with  dry  rubble  or  with  cut  stone 
in  hydraulic  mortar,  great  attention  should  be  paid  to  details 
of  workmanship  and  construction,  all  of  which  should  be  clearly 
set  forth  in  the  specifications.  The  earth  must  be  thrown 
loosely  against  the  wall  or  be  dumped  against  it  from  above, 
but  should  be  carefully  packed  in  layers  which  slope  upward 
toward  the  back. 

Problem  27.  Let  0  =  38  degrees,  6—  10  degrees,  w=  100 
pounds  per  cubic  foot,  £>:=  150  pounds  per  cubic  foot,  #—2 
feet,  Q  —  80  degrees.  Compare  the  quantities  of  material  re- 
quired for  two  walls,  one  9  feet  high  and  the  other  18  feet 
high. 


ART.  28.]  THE  PRESSURE  OF   WA  TER.  93 


CHAPTER  V. 
MASONRY    DAMS. 

ARTICLE  28.    THE  PRESSURE  OF  WATER. 

All  doubts  regarding  the  direction  and  intensity  of  the 
lateral  pressure  against  walls  vanish  when  the  earth  is  replaced 
by  water.  For  since  water  has  no  angle  of  repose,  0  =  o°  and 
6  =  0°,  and  all  the  formulas  of  Chapter  II  reduce  to  (35), 
which  gives  the  normal  water  pressure  against  a  wall  of  height 
h  when  the  depth  of  the  water  is  also  h. 

The  principles  of  hydrostatics  show  that  the  direction  of 
water  pressure  is  always  normal  to  a  submerged  plane  ;  also 
that  the  total  normal  pressure  on  such  a  surface  is  obtained  by 
multiplying  together  the  weight  of  a  cubic  unit  of  water,  the 
area  of  the  surface  and  the  depth  of  its  centre  of  gravity  below 
the  water  level. 

The  water  level  is  usually  lower  than  the  top  of  the  dam, 
as  shown  in  Figure  27.  Let  d  be  the  vertical  depth  of  the  water 
above  the  base  of  a  trapezoidal  dam,  6  the  angle  which  the 
back  makes  with  the  horizontal,  and  w  the  weight  of  a  cubic 


94  MASONRY  DAMS.  [CHAP.  V. 

unit  of  water.     Then  the  surface  submerged  is  — — -»  X  I,  the 

depth  of  its  centre  of  gravity  below  the  water  level  is  \dt  and 
hence  the  normal  pressure  is 

P=\wd*  -v-  sin  0, 


which  agrees  with  (35).  The  centre  of  pressure,  or  the  point 
at  which  the  resultant  pressure  must  be  applied  to  balance  the 
actual  presssures,  is  on  the  back  of  the  dam  at  a  vertical  height 


FIG.  27. 

of  \d  above  the  base  ;  this  is  known  by  a  theorem  of  hydro 
statics  and  likewise  by  Article  12. 

The  angle  6  is  never  less  than  a  right  angle  for  masonry 
dams,  and  hence  it  will  be  convenient  to  use  instead  of  it  the 
angle  ip  which  the  plane  of  the  back  makes  with  the  vertical. 
Then  6  =  90°  -f-  ip,  and  the  normal  pressure  is 


P  =  $wda  sec  i/>,     ......     (69; 

>jue,  $f~ 

and  for  a  vertical  wall,  where  i/>  =  o°,  this  becomes  P= 


ART.  29.]  PRINCIPLES  AND  METHODS,  95 

The  normal  pressure  P  may  be  decomposed  into  a  hori- 
zontal component  P  '  and  a  vertical  component  P  "  ,  whose  values 
are  expressed  by 


P1  =  Pcos  $  =  %wd\       P"  =  Psin  $  =  %u>d*  tan  ^  ;         (70) 

and  if  ^  be  a  small  angle,  as  is  usually  the  case,  the  horizon- 
tal component  \wd*  is  sometimes  taken  as  the  actual  water 
pressure.  This  is  an  error  on  the  side  of  safety,  since  the  ver- 
tical component,  acting  downward,  increases  the  stability  of 
the  dam,  unless  the  water  penetrates  under  the  base  BC,  which 
is  an  element  of  danger  that  ought  not  to  be  allowed. 

Problem  28.  For  a  waste-weir  dam  the  water  level  may  be 
higher  than  AD  by  an  amount  d,.  Prove  that  the  normal 
pressure  is 

-cosifr,      ....    (71) 


and  that  the  centre  of  pressure  is  at  a  vertical  distance^  above 
B,  whose  value  is  given  by  the  formula 

h  d 


ARTICLE  29.    PRINCIPLES  AND  METHODS. 

The  fundamental  requirements  concerning  the  design  of 
masonry  dams  are  the  same  as  those  governing  all  engineering 
-work  ;  first,  stability,  and  second,  economy.  The  first  requires 
that  the  structure  be  built  so  that  all  its  parts  shall  have 


g6  MASONRY  DAMS.  [CHAP.  V. 

proper  strength,  and  the  second  that  this  shall  be  done  with 
the  least  total  expenditure  of  money.  This  expenditure  con- 
sists of  two  parts,  that  for  material  and  that  for  labor,  and 
economy  will  result  if  material  can  be  saved  without  increasing 
the  labor.  Hence  all  parts  of  a  structure  ought  to  be  of 
equal  strength  (like  the  "  one-hoss  shay  "),  provided  that  the 
cost  of  the  material  thus  saved  is  greater  than  the  cost  of  the 
extra  labor  required ;  for  if  one  part  exceeds  the  others 
in  strength  it  has  an  excess  of  material  which  might  have 
been  saved. 

For  ordinary  retaining  walls  and  for  low  masonry  dams  the 
trapezoidal  form  is  the  only  practicable  cross-section,  since 
curved  faces  do  not  save  sufficient  material  to  balance  the  cost 
of  the  extra  expense  of  construction.  But  for  high  masonry 
dams,  and  as  such  may  be  classed  those  over  80  or  100  feet 
high,  it  not  only  pays  to  deviate  from  the  trapezoidal  section, 
but  it  is  often  absolutely  necessary  to  do  so  in  order  to  reduce 
the  pressure  on  the  base  to  allowable  limits.  The  section 
adopted  in  such  cases  is  therefore  an  approximation  to  that  of 
a  form  of  uniform  strength. 

The  general  principles  of  stability  of  retaining  walls  set 
forth  in  the  preceding  pages  apply  to  all  masonry  structures, 
but  it  will  be  well  to  state  them  briefly  again,  with  especial 
reference  to  dams. 

First,  there  must  be  proper  stability  against  sliding  at 
every  joint  and  at  every  imaginary  horizontal  section.  This 
can  be  done  either  by  bonding  the  masonry  with  random 


ART.  29.]  PRINCIPLES  AND   METHODS.  97 

courses  so  that  no  through  joints  exist,  or  by  inclining  such 
joints  at  the  proper  backward  slope  (Article  23).  The  first 
method  is  alone  applicable  to  a  dam,  and  by  the  use  of 
hydraulic  mortar  the  whole  structure  should  be  made 
monolithic. 

Second,  there  must  be  proper  stability  against  rotation  at 
every  horizontal  section  of  the  dam.  This  will  be  secured 
when  the  resultant  of  all  the  forces  above  that  imaginary  base 
cuts  it  within  the  middle  third  (Article  17)  or  at  the  most  at 
the  limit  of  the  middle  third.  In  a  dam  there  will  be  two 
cases  to  be  considered :  (a)  when  the  reservoir  is  full  of  water, 
and  (£)  when  the  reservoir  is  empty.  For  the  first  case  the 
line  of  resistance  should  not  pass  without  the  middle  third  on 
the  front  or  down-stream  side,  and  for  the  second  case  it  should 
not  pass  without  it  on  the  back  or  up-stream  side. 

Third,  there  must  be  proper  security  against  crushing  at 
every  point  within  the  masonry.  As  a  general  rule  this  de- 
mands that  the  compressive  stress  per  square  inch  shall  not 
exceed  150  pounds,  although  in  a  few  cases  higher  values  have 
been  allowed. 

It  will  be  found  in  designing  a  high  dam  that  the  second 
principle  will  determine  the  thicknesses  for  about  100  feet 
below  the  top.  For  greater  heights  the  third  principle  must 
generally  be  used,  and  the  formulas  of  Article  20  be  applied. 
It  is  indeed  doubtful  whether  these  formulas  correctly  repre- 
sent the  actual  distribution  of  stress  on  the  base  of  a  high 
dam  with  a  polygonal  cross-section,  for  it  would  naturally  be 


98  MASONRY  DAMS,  [CHAP.  V. 

thought  that  greater  stresses  would  obtain  near  the  middle 
rather  than  near  the  edge  of  the  base.  If  such  is  the  case, 
however,  the  application  of  the  formulas  can  only  err  on  the 
side  of  safety. 

Problem  29.  A  masonry  dam  36  feet  high  and  24  feet 
wide  weighs  150  pounds  per  cubic  foot.  Find  the  point 
where  the  resultant  cuts  the  base  when  the  water  is  33  feet 
deep  above  the  base. 


ARTICLE  30.    INVESTIGATION  OF  A  TRAPEZOIDAL  DAM. 

1  6i      ?4,*C  *^  (^^^^CL^j^ 

The  given  data  will  furnish  the  dimensions  of  the  dam,  and 

the  normal  water  pressure  on  its  back  will  be  computed  by 
(69).  Then  by  the  method  of  Article  18  a  graphical  investiga- 
tion for  rotation  may  be  made  and  the  factor  of  security  be 
determined  for  any  joint  BC.  Through  joints  should  not 
exist  in  a  masonry  dam,  and  hence  BC  will  be  taken  as  hori- 
zontal in  the  construction,  or  even  if  they  do  exist  BC  may  be 
an  imaginary  horizontal  joint. 

The  factor  of  security  against  rotation  may  be  computed 
by  the  formulas  of  Article  19,  first  making  z  =  o°,  and  9  = 
90°  +  ^.  Then  from  the  given  datq^P  is  found  by  (69),  Fand 
Vs  by  (51)  and  (52),  and  /  is  computed  by  (50),  in  which  h  is 
to  be  put  equal  to  */,  whence  finally  n  is  derived  by  (48).  It 
will  however  be  more  satisfactory  for  a  student  to  make  an 
analysis  directly  from  first  principles  rather  than  to  arbitrarily 
use  formulas  for  mere  computation.  This  will  be  now  done 
for  a  particular  example. 


ART.  30.]  INVESTIGATION  OF  A    TRAPEZOIDAL  DAM, 


99 


The  largest  trapezoidal  dam  is  that  at  San  Mateo,  CalK 
fornia.  The  top  thickness  is  20  feet,  the  base  thickness  is  176 
feet,  the  vertical  height  is  ijh  feet,  the  batter  of  the  back  is  i 
to  4,  and  the  masonry  is  concrete,  which  probably  weighs  about 
150  pounds  per  cubic  foot.  It  is  required  to  investigate  itc 
stability  when  the  water  is  165  feet  deep  above  the  base. 

Let  P  be  the  normal  water  pressure  on  the  back,  and  V 
the  weight  of  dam,  both  per  foot  of  length.  Let  BC  be  the 
base,  and  T  the  point  where  the  resultant  of  P  and  V  cuts  it. 

.   & 

,D 


C 


FIG.  28. 


f(/U*^ 
Qt 

Now  with  respect  to  this  point  the  moment  of  Pwill  equal 
the  moment  of  V.-  Let/  be  the  lever-arm  of  P\  let  /  repre- 
sent the  distance  CT,  and  s  the  horizontal  distance  from  B  to 
the  line  of  direction  of  V.  The  lever-arm  of  V  is  then 
b  —  s  —  /,  and  the  equation  of  moments  is 


Pp  =  V(b-s-  f)    or 


—  Vs—  Vt. 


(73) 


The  first  member  of  this  equation  may  be  replaced  by 
Pp'  —  P"p",  in  which  P'  and  P"  are  the  horizontal  and  verti- 
cal components  of  P,  and  p'  and  p"  are  their  lever-arms  with 


IOO  MASONRY  DAMS.  [CHAP.  V. 

respect  to  the  point  T.  Also  V  may  be  replaced  by  vAr 
where  v  denotes  the  weight  of  the  masonry  per  cubic  foot  and 
A  is  the  area  of  the  cross-section.  Then 

P'p'  -P"p"=.v(Ab-As-At\     .     .     .     (74) 
which  is  a  formula  better  adapted  to  numerical  operations. 

To  apply  this  to  the  San  Mateo  dam  the  data  are  d  =  165 
feet,  tan  0  =  0.25,  a  =  20  feet,  b  =  176  feet,  h  =  170  feet,  and 
v  =  150  pounds  per  cubic  foot.  Then  from  (70) 

P'  =  850  780  pounds,     P"  =  o.2$P  =  212  700  pounds, 

4  r  4-U     G 

and  from  the  figure, 

/  =  &  =  55  feet,        /'  =  176  -  0.25  X  55  -  '• 
Also  the  area  of  the  trapezoid  is 

A  =  i  X  170(176  -f-  20)  =  1 6  660  square  feet, 

and  the  moment  As  is  computed  by  regarding  A  as  the  sum  of 
the  triangles  AaB  and  DdC  and  the  rectangle  AadD  (Figure 
28),  thus  ;  ^  /^P^&ttf  ^C*/  >^UrZ^L4^f>>^ 

6_  »  t>~  t  UK  <  :'.  r  i  ^Jfjy  &  AA+**++/  ty  4stsL4Usvi*C't*£t 

As=AaBx  %Ba  +  AadD(Ba  +  Ja^jf  +  DdC(BC  —  \dC\ 

t^i^^^f-i  cAc*~^.^-i^  /^-^-i^wc^rr 

whence  As  =  I  248  820  feet  cube.  Inserting  now  all  values  in 
(74)  and  solving  for  t  there  is  found  /  =  88.6  feet.  The  result- 
ant therefore  cuts  the  base  very  near  the  middle,  so  that 
the  factor  of  security  against  rotation  is  practically  infinity 
(Article  17). 


ART.  30.]   INVESTIGATION  OF  A    TRAPEZOIDAL  DAM.  IOI 

It  is  the  custom  of  some  engineers  to  neglect  the  vertical 
component  of  the  water  pressure,  and  regard  only  the  horizontal 
component.  Testing  the  San  Mateo  dam  under  this  supposi- 
tion, P"  equals  zero,  and,  all  other  quantities  being  the  same 
as  before,  there  is  found  t  =  80.2  feet,  whence  the  factor  of 
security  against  rotation  is 

88 


which  shows  that  the  degree  of  stability  is  ample. 

A  masonry  dam  should  be  investigated  not  only  for  the 
case  when  the  reservoir  is  filled  with  water,  but  also  for  the 
case  when  the  reservoir  is  emptied.  Here  the  tendency  to 
rotation,  or  overturning,  is  usually  backward  instead  of  for- 
ward. Let  5  be  the  point  where  the  direction  of  the  weight  V 
cuts  the  base,  and  let  M  be  the  middle  of  the  base.  Then  the 
factor  of  security  is  the  ratio  of  MB  to  MS,  or 


in  which  the  distance  s  is  computed  by  dividing  the  value  of 
As  by  that  of  A.     Now,  for  the  San  Mateo  dam, 

I  248  820 

:  — J— —  =  75.0  feet, 
16660 

and  then  n  is  found  to  have  a  value  of  6.8. 


IO2  MASONRY  DAMS.  [CHAP.  V. 

No  through  joints  exist  in  this  dam,  and  the  method  of 
construction  of  the  base  is  such  as  to  preclude  all  possibility  of 
sliding.  Moreover  by  the  use  of  (43)  the  coefficient  of  friction 
which  will  allow  sliding  to  occur  on  the  base  is 

850780 
~~  I50XI6660"0'34' 

a  value  which  would  be  very  low  for  an  imperfect  construction. 

The  compressive  stresses  on  the  base  may  next  be  investi- 
gated by  the  method  of  Article  20.  When  the  water  in  the 
reservoir  is  165  feet  deep  the  resultant  R  cuts  the  base  so  near 
the  middle  that  the  compression  can  be  regarded  as  uniformly 
distributed.  The  pressure  normal  to  the  base  is  V-\-  P"t  and 
hence  the  stress  per  square  inch  is 

150  X  16660  +  212700 

S  = -2 =  107  pounds, 

144  X  176 

which  is  probably  less  than  one-sixteenth  of  the  ultimate 
strength  of  good  concrete  when  one  year  old. 

If  the  reservoir  should  be  empty  the  greatest  stress  would 
come  at  the  heel  B,  and  as  V  is  applied  at  75  feet  from  B,  that 
stress  in  pounds  per  square  inch  is,  from  formula  (58), 

2X  150  X  16660  /         3  X  75\ 

O  ^^  — ~^ —  "  I  •£  —  ~    — ? —  j  ==   14.2. 

144  X  176  176   / 

It  will  also  be  found  that  the  stress  at  the  middle  of  the  base  is 
99  pounds  per  square  inch,  that  at  the  toe  C  is  142  —  99  =  43 
pounds  per  square  inch. 


ART.  si.]  DESIGN  OF  A   LOW    TRAPEZOIDAL  SECTION.  IO3 

Problem  30.  Investigate  the  security  of  the  San  Mateo 
dam  for  a  horizontal  section  100  feet  below  its  top — (a)  when 
the  water  is  95  feet  deep  above  that  section  ;  (b)  when  the 
reservoir  is  empty. 


ARTICLE  31.    DESIGN  OF  A  Low  TRAPEZOIDAL  SECTION. 

When  a  trapezoidal  dam  is  to  be  designed  its  height  h  will 
be  given,  and  also  the  depth  d  of  the  water  behind  it.  The 
weight  per  cubic  foot  of  the  masonry  v  will  be  known,  at  least 
approximately.  The  thickness  of  the  top,  #,  will  be  assumed ; 
usually  this  will  serve  for  a  roadway  or  footway  and  hence 
cannot  be  less  than  8  or  10  feet.  The  batter  of  the  back,  or 


tan  0,  is  next  assumed,  and  usually  this  will  be  taken  small  in 
order  that  the  weight  of  the  wall  V  may  fall  as  far  away  from 
the  toe  C  as  possible.  Let  M  be  the  middle  of  the  base  BC ;  let 
5  be  the  point  where  the  direction  of  Fcuts  it,  and  T^the  point 
where  the  direction  of  the  resultant  R  cuts  it.  It  is  plain  that 
MS  will  always  be  less  than  one-third  of  MB  for  any  trapezoid 
whose  back  leans  forward,  and  that  it  becomes  equal  to  one- 
third  of  MB  only  when  AD  is  zero  and  AB  is  vertical. 


' ' 


IO4 


MASONRY  DAMS. 


[CHAP.  V. 


Let  b  be  the  length  of  the  base  BC,  and  let  t  be  the  distance 
CT.  It  is  required  to  findj£  so  that  MT  shall  be  one-third  of 
MC,  or,  what  is  the  same  thing,  that  t  shall  equal  %b.  Full 
security  against  rotation  will  then  exist  both  for  reservoir  full 


Formula  (74)  is  a  funamental  one  applicable  to  any  sec- 
tion.    To  aoply  it  tathe  problem  in  hand,  the  values  of  the 
lever-armsandjg'  are  to  be  stated  in  terms  of  the  otter    y— 
s  :  fcd*Jv       **s*<<<* 


quantities,  thus 


-.        .     . 

&£*ZCLvr  jjL    ^^p^^^^uL^  '-t&t-c 

Also  the  area  A  is  expressed  by   ) 

'  0Kt$M6&  J 

.    .....     (76) 


and    by   the    method    of   the   last   Article   the  value   of   the 
moment  As  is  found  to  be 


.     (77) 


As  =        u"  +  ab  +  P  +  h(2a  +  b)  tan 


Inserting,  now,  all  these  quantities  in  (74),  and  making  /  =  %b, 
there  is  found  a  quadratic  equation  in  b  whose  solution  gives 


b—  -F+VF' 
in  which  F  and  G  have  the  values 


(78) 


G  =  2--(P'  +  P"  tan  0)  +  a*  +  2ah  tan  t/>, 


ART.  31.]  DESIGN  OF  A    LOW   TRAPEZOIDAL   SECTION.  IO5 

and  from  these  the  proper  base  thickness  can  be  found,  P'  and 
P"  being  first  computed  by  (70),  or  if  desired  the  expressions 
for  F  and  G  can  be  written 


tan*; 


gh 

in  which  g  is  the  ratio  of  v  to  w,  or  the  specific  gravity  of  the 
masonry. 

If  ty  =  o°,  the  formula  (78)  takes  the  simple  form 


•    •    •    •    (79) 

which  gives  the  proper  base  thickness  of  a  trapezoidal  dam 
with  a  vertical  back. 

~f  o     --g: 

The  compressive  stress  at  C  in  pounds  per  square  inch  is 
now  found  from  (5*5),  or 

//  'an». 


144^ 

and  if  this  is  less  than  the  specified  limiting  value,  no  further 
investigation  will  be  necessary  ;  but  if  greater,  then  the  above 
formulas  for  thickness  will  not  apply  and  those  of  the  next  Ar- 
ticle must  be  used.  The  limiting  value  of  5  is  often  taken 
at  150  pounds  per  square  inch. 

The  compression  at  the  inner  edge  B  when  the  reservoir  is 
empty  is  less  than  that  at  C  when  it  is  full,  for  in  any  trape- 


IO6  MASONRY  DAMS.  [CHAP.  V. 

zoid  where  tan  fy  is  positive  MS  is  less  than  one-third  of  MB. 
Th€  distance  BS  can,  however,  be  obtained  by  dividing  (77) 
by  (76),  whence 


_ 

"*HhflT 

and  then  by  the  use  of  (58)  the  unit-stress  at  .#  is  computed. 

In  order  to  show  the  application  of  the  formulas  and  at  the 
same  time  study  the  question  of  economic  proportions,  let  the 
following  data  be  taken  :  h  =  60  feet,  d  =  57  feet,  a  —  9  feet, 
v  =  150  pounds  per  cubic  foot  or^-=  2.4.  Let  three  designs 
be  made  for  which  the  back  has  different  batters,  namely, 
tan  #  =  -J-,  tan  $  =  -fa,  and  tan  $  —  o.  Using  the  formula  (78), 
the  base  is  first  found,  and  then  by  (76)  the  area  of  each 
trapezoid  ;  thus  : 

tan  i/>  =  -J-,  b  =  36.5  feet,  A  —  1365  sq.  ft.,  =  109  per  cent 
tan  i/>=^,  b  =  34.4  feet,  A  =  1302  sq.  ft.,  =  104  per  cent 
tan  i/}  =  o,  b  —  32.75  feet,  A  =  1253  sq.  ft.,  =  100  per  cent 

From  which  it  is  seen  that  the  most  advantageous  section  is 
the  one  with  the  vertical  back.  This  conclusion  might  also  be 
inferred  from  the  discussion  in  Article  24. 

It  is  the  custom  of  some  engineers  to  neglect  the  vertical 
component  of  the  water  pressure.  Formula  (78)  may  be 
adapted  to  this  hypothesis  by  making  P"  equal  to  zero  in  the 
quantities  F  and  G,  which  then  become 


73 

G  =  ---  1-  c?  +  2ak  tan  ib. 
gh 


ART.  32.]  DESIGN  OF  A   HIGH  TRAPEZOIDAL   SECTION.         IO? 

The  thickness  of  the  dam  computed  under  this  hypothesis  is 
greater  than  before.  Thus,  for  the  above  example, 

tan  tp  =  |,  b  =  39.8  feet,  A  —  1466  sq.  ft.,  =117  per  cent 
tan  $  =:  1ig.,  b  =  36.2  feet,  A  —  1356  sq.  ft,  =  108  per  cent 
tan  ^  —  o,  b  —  32.75  feet,  A  =  1253  sq.  ft,  =  100  per  cent 

Problem  31.  Find  the  compressive  unit-stress  at  B  and  C  for 
one  of  the  cases  of  the  above  numerical  example. 

ARTICLE  32.  DESIGN  OF  A  HIGH  TRAPEZOIDAL  SECTION. 

When  the  value  of  h  is  so  great  that  the  formula  for  thick- 
ness deduced  in  the  last  article  cannot  be  used  the  dam  is  said 
to  be  "  high."  For  such  cases  the  condition  /  =  \b  cannot  be 
applied,  but  /  must  be  made  greater  than  \b  so  as  to  reduce 
the  unit-stress  at  the  toe  C.  The  base  thickness  will  hence  be 
greater  than  that  given  by  (78). 

Let  5  be  the  given  limiting  unit-stress  in  pounds  per  square 
foot.  The  corresponding  value  of  /  is,  from  (57),^s.  Q  „ 

Trr 


in  which  vA  is  the  equivalent  of  the  weight  V.  Inserting  this 
in  (74),  and  also  the  values  for  / ',  /  " ,  A  and  As,  there  is  de- 
duced a  quadratic  in  b  whose  solution  gives 


108  MASONRY  DAMS.  [CHAP.  V. 

in  which  K  and  Z,  have  the  values 
K=(P"  —  i^2tan^)-i 


"  tan  $)  +  2/%2  +  2^  tan  #). 

If  in  these  P"  =  o,  the  vertical  component  of  the  water  press- 
ure is  neglected  ;  and  if  tan  fi  =  o,  the  back  of  the  trapezoid  is 
vertical. 

In  using  these  formulas  the  given  data  are  a,  k,  d,  tan  ^,  v 
and  5.  Then  b  is  computed,  taking  the  water  pressures  P'  and 
P"  from  (75).  When  b  is  found,  s  should  be  determined  by 
(81),  and  then  by  (58)  the  stress  at  B  when  the  reservoir  is 
empty  is  computed. 

For  an  example  take  a  —  20  feet,  h  =  170  feet,  d=  165 
feet,  tan  i(>  =  o.2,v=  1  50  pounds  per  cubic  foot  and  5  =  2  1  ooo 
pounds  per  cubic  foot.  Let  it  be  required  to  find  b,  neglect- 
ing the  vertical  component  P"  of  the  water  pressure.  From 
Article  28  the  value  of  P'  is  850  780  pounds,  and  by  hypothesis 
P"  —  o.  Then  inserting  all  values,  K  —  —  20.64,  L  =  15  506.6, 
whence  b  =  145.2  feet.  This  gives  for  the  area  of  the  section 
A  —  14  042  square  feet,  and  from  (82)  t  •=.  0.425  b,  which  locates 
the  point  where  the  resultant  pierces  the  base  when  the  reser- 
voir is  full.  From  (81)  there  is  found  s  =  61.9  feet  =  0.426$, 
which  gives  the  point  where  the  line  of  action  of  Fcuts  the 
base,  and  when  the  reservoir  is  empty  the  unit-stress  at  the 
.back  edge  of  the  base  is,  by  (58), 

5,  =  ^  150X14042  (2  _  3  x 

145.2 


ART.  33.]        ECONOMIC  SECTIONS  FOR  HIGH  DAMS.  IOO, 

so  that  the  compression  at  B  for  reservoir  empty  is  about  the 
same  as  that  at  C  for  reservoir  full. 

Problem  32.  Discuss  the  above  example  without  neglecting 
the  vertical  component  of  the  water  pressure. 


ARTICLE  33.    ECONOMIC  SECTIONS  FOR  HIGH  DAMS. 

A  high  trapezoidal  dam  designed  so  as  to  give  proper  se- 
curity against  crushing  on  the  base  has  an  excess  of  stability 
in  its  upper  part.  Accordingly  if  the  section  be  polygonal,  or 
bounded  by  curved  lines,  both  in  front  and  back,  these  may 
be  arranged  so  as  to  save  material  in  the  upper  parts,  thus  less- 
ening the  weight  that  comes  on  the  base,  and  hence  reducing 
its  width  from  that  which  a  trapezoidal  section  would  require. 
Such  a  structure  will  be  approximately  one  of  uniform  security 
against  rotation  in  its  upper  portions,  and  of  uniform  security 
against  crushing  in  its  lower  portions.  The  method  of  design 
ing  the  upper  part  will  be  similar  to  that  used  in  Article  26  for 
the  retaining  wall. 

Local  and  practical  considerations  will  determine  the  thick- 
ness of  the  top  AD.  From  the  principles  deduced  in  Articles 
24  and  31  it  is  plain  that  to  secure  the  greatest  economy  of 
material  the  back  should  be  vertical  for  some  distance  below 
the  top.  If  the  upper  sub-section  AA'D'D  be  rectangular,  the 
line  of  resistance  for  the  case  of  reservoir  empty  will  cut  the 
middle  of  A'D' ;  and  if  the  height  be  properly  chosen,  the  line 
of  resistance  for  reservoir  full  will  cut  it  at  the  front  edge  of 


no 


MASONRY  DAMS. 


[CHAP.  V. 


the  middle  third.     To  find  what  this  height  should  be  let  a  be 
the  thickness,  h'  the  height  A  A ',  and  d  the  depth  of  water 


FIG.  30. 


above  A'.     Then  the  equation  of  moments  with  reference  to  a 
point  in  the  base  distant  \a  from  D'  is 

%wd*  X^  =  vati  X  \a. 

Now  if  d  be  taken  equal  to  h',  as  it  may  be  in  an  extreme  case, 
the  solution  of  this  gives 


(84) 


in  which  g  is  the  ratio  of  v  to  wy  or  the  specific  gravity  of  the 
masonry. 

The  next  sub-section  should  be  a  trapezoid,  and  the  entire 
section  in  fact  may  be  considered  as  made  up  of  trapezoids, 
the  widths  of  these  being  so  determined  as  to  secure  economy 
and  stability.  The  former  requires  that  the  back  should  be 
vertical  or  that  its  batter  should  be  as  small  as  possible,  and 


" 


fa 


ART,  33.]        ECONOMIC  SECTIONS  FOR  HIGH  DAMS.  Ill 

the  latter  requires  that  the  lines  of  resistance  for  reservoir  full 
and  reservoir  empty  shall  not  pass  without  the  middle  third, 
while  the  resulting  unit-stresses  are  kept  within  the  specified 
limit. 

In  the  upper  part  of  the  dam  the  question  of  the  com- 
pression of  the  masonry  need  not  be  considered,  and  the  width 
of  the  base  of  each  sub-section  will  be  found  from  the  require- 
ment that  the  line  of  resistance  for  reservoir  full  cuts  that  base 
at  one-third  the  length  from  the  front  edge. 

In  the  lower  part  of  the  dam  the  widths  are  to  be  de- 
termined by  regarding  the  compressive  stresses.  Owing  to 
uncertainties  concerning  the  theor)^  of  distribution  of  these 
stresses,  and  to  differences  of  opinion  concerning  the  manner 
in  which  it  should  be  applied,  engineers  have  not  agreed  upon 
a  uniform  method  of  design.  The  general  form  of  section, 
however,  is  that  shown  in  Figure  30,  the  back  being  battered 
below  a  certain  depth  in  order  to  keep  the  line  of  resistance 
for  reservoir  empty  well  within  each  base,  while  the  batter  of 
the  front  increases  downward.  The  views  of  different  authori- 
ties are  fully  set  forth  in  WEGMANN's  Design  and  Construc- 
tion of  Masonry  Dams  (second  edition,  New  York,  1889), 
where  also  are  given  sectional  drawings  of  all  existing  high 
•dams. 

Problem  33.  Prove  that  a  triangular  section  is  one  of  uni- 
form stability  against  rotation  when  the  water  level  is  at  the 
vertex  of  the  triangle. 


112  MASONRY  DAMS,  [CHAP.  V. 


ARTICLE  34.    INVESTIGATION  OF  A  POLYGONAL  SECTION. 

The  graphical  investigation  of  the  stability  and  security  of 
a  polygonal  section  like  Figure  30  is  so  simple  in  theory  that 
space  need  not  here  be  taken  to  set  it  forth  in  detail.  The 
general  method  of  Article  18  is  to  be  followed  for  the  base  of 
each  sub-section,  and  the  only  difficulty  that  need  to  occur  will 
be  in  connection  with  determining  the  positions  of  the  centres 
of  gravity  of  the  areas  above  the  successive  bases.  These  may 
be  best  computed  by  the  method  explained  bdow.  When  the 
points  5  and  T  have  been  found  for  each  base  the  factor  of 
security  against  rotation  is  known  by  Article  17,  both  for  reser- 
voir full  and  reservoir  empty,  and  then  the  maximum  com- 
pressiye  stresses  are  determined  as  in  Article  20. 


IV    ^    "*  ^T^K^tZ 

The  analytical  investigation  begins  with  the  top  sub-section, 

which  is  either  a  rectangle  or  a  trapezoid  (Figure  30),  and  finds 
as  in  Article  30,  or  by  the  formulas  of  Article  19,  the  degree  of 
security  for  its  base  A'D'.  Thus  is  determined  the  area  A^ 
the  corresponding  weight  V^  and  the  horizontal  distance  sl 
from  its  point  of  application  to  its  back  edge.  Now  let 
A' BCD'  be  the  next  trapezoid,  let  h  be  its  vertical  height,  a 
its  top  width,  b  its  base  width,  ip  the  angle  of  inclination  of  the 
back  to  the  vertical,  A^  its  area,  v  the  weight  of  the  masonry 
per  cubic  unit,  V^  its  weight  vA9  which  is  applied  at  a  horizon- 
tal distance  52  from  the  back  edge  B.  The  sum  Al-\-A9  is 


ART.  34.]  INVESTIGATION  OF  A    POLYGONAL   SECTION.         113 

the  total  area  A  whose  weight  is  vA  =  F,  and  the  line  of  action 
of  this  cuts  the  base  at  S,  whose  horizontal  distance  from  the 


back  edge  B  is  called  s.     The  value  of  s  can  be  obtained  by 
taking  moments  about  B,  thus  : 


which  is  the  formula  for  locating  the  line  of  resistance  when 
the  reservoir  is  empty.     The  values  of  A^  and  A2s^  are  found/'' 
from  the  given  quantities  a,  b,  //,  tan  fy  by  the  help  of   (76) 

and  (77).  ^  =-  -fcfcfi  V^£*$  VtfA«**^2 

When  the  reservoir  is  full  let  Pf  be  the  horizontal  compo- 
nent of  the  water  pressure  on  the  entire  back  above  B,  and  P" 
the  vertical  component.  Let  their  lever-arms  with  respect  to 
Tbep'  and/".  Then  the  equation  of  moments  is 

P'p'  -P"p"^v(A^A^(b^-s-t\     .     .     (86) 

In  this  the  value  of  P'  is  %wd*,  and  that  of  p'  is  %d.  If  the 
batter  of  the  back  be  uniform  from  the  top  to  B,  the  values  of 
P"  and  p"  are  known  by  (70)  and  (75).  If,  however,  the  dif- 
ferent trapezoids  have  different  batters,  values  for  P"  and  p" 


are  not  easily  expressed.     Hence  it  is  often  customary  to  neg- 
lect P",  and  then  the  distance  CT  is 

......     (87) 


in  which  g  is  the  specific  gravity  of  the  masonry.     From  this 
the  line  of  resistance  can  be  located  when  the  reservoir  is  full. 

The  factor  of  security  against  rotation  can  now  be  found, 
if  desired,  by  (45)  both  for  the  case  of  reservoir  empty  and 
that  of  reservoir  full.  The  degree  of  security  against  crushing 
will  be  deduced  by  computing  the  unit-stresses  at  B  and  £7  by 
the  help  of  the  formulas  of  Article  20  and  then  comparing 
these  with  allowable  and  with  ultimate  values.  The  degree 
of  security  against  sliding  could  be  easily  determined  if  the 
coefficient  of  friction  were  known,  but  as  the  base  is  not  a 
real  joint,  it  will  be  sufficient  to  use  formula  (39),  and  deduce 
the  value  of  f  which  would  allow  motion  if  a  joint  actually 
existed. 

The  above  formulas  can  be  applied  to  each  trapezoid  in 
succession,  Al  being  taken  as  all  the  area  above  its  top,  and 
thus  the  lines  of  resistance  can  be  traced  throughout  the  en- 
tire section. 

As  a  numerical  example  let  it  be  required  to  test  the  fourth 
trapezoid  of  the  theoretical  section  of  the  Quaker  Bridge  Dam 
given  in  Article  35.  The  data  are  Al  =  1823  square  feet,  sl  = 
12.4  feet,  h  =  20  feet,  tan  ^  =  0.115,  a  =  37.4  feet,  b  =  53.4 


ART.  35.]      DESIGN  OF  A   HIGH  ECONOMIC  SECTION.  115 

feet,  d  =  go  feet,  and  g—  —  =  2 J-;  and  it  is  required  to  find  s 

w 

and  /  with  the  unit-stresses  5,  and  5.  First  the  area  of  the 
given  trapezoid  is  908  square  feet,  and  its  moment  A^s^is 
21  808  feet  cube.  Then  from  (85)  the  value  of  s  is  17.8  feet, 
and  inserting  this  in  (87)  there  is  found  t  —  17.8  feet.  The 
lines  of  resistance  here  cut  the  base  at  the  ends  of  the  middle 
third  so  that  the  factors  of  security  for  reservoir  full  and  for 
reservoir  empty  are  each  3.0  (Article  17).  The  unit-stresses 
^  and  5  are  also  equal,  and  each  will  be  found  to  be  1 1 1 
pounds  per  square  inch.  Lastly,  from  (i)  or  (39)  the  coeffi- 
cient of  friction  necessary  for  equilibrium  is  0.59,  a  value 
which  cannot  be  approached  in  a  monolithic  structure. 

Some  authors  use  the  term  "  factor  of  safety  "  as  meaning 
the  ratio  of  the  horizontal  water  pressure  which  would  cause 
overturning  to  the  actual  existing  horizontal  water  pressure. 
This  should  not  be  confounded  with  the  factor  of  security 
used  in  this  book. 

Problem  34.  Given  #,  #,  h,  Vl  and  sl  for  any  trapezoid 
(Figure  31).  Deduce  the  value  of  tan  ^  so  that  s  shall 
equal  %b. 

ARTICLE  35.    DESIGN  OF  A  HIGH  ECONOMIC  SECTION. 

The  application  of  formula  (86)  will  in  general  lead  to  com- 
plicated equations,  unless  the  vertical  component  of  the  water 
pressure  P"  is  neglected.  This  is  an  error  on  the  side  of  safety 
and  is  hence  often  allowable,  particularly  when  tan  $  is  small. 


Il6  MASONRY  DAMS.  [CHAP.  V. 

The  following  method  is  essentially  like  that  devised  by 
WEGMANN  for  the  design  of  the  Quaker  Bridge  Dam,  and  is 
here  given  because  of  all  the  different  methods  it  appears  to 
be  best  adapted  to  the  comprehension  of  students. 

Using  the  same  notation  as  in  the  last  article,  the  top  width 
a  is  first  assumed,  and  the  uppermost  sub-section  is  made  a 
rectangle  whose  height  is  by  (84)  equal  to  a  Vg.  The  follow- 
ing  sub-sections  will  be  trapezoids  with  vertical  backs,  each 
base  being  determined  so  that  /  =  ^b.  To  find  b  for  any 
trapezoid  there  will  be  given  At  and  sl  from  the  preceding 
trapezoids,  its  upper  base  a,  its  height  h,  the  total  depth  of 
water  d,  and  the  specific  gravity  g,  while  tan  ip  equals  zero. 
First  A^  and  A^  are  expressed  in  terms  of  a,  b  and  ^,  by  (76) 
and  (77),  and  these  are  inserted  in  (85).  Then  the  resulting 
expression  for  s  is  put  into  (86)  and  /  made  equal  to  \b.  Thus 
is  obtained  a  quadratic,  whose  solution  gives 

b—  -K+VK*+L,  .....  (88) 

in  which  K  and  L  have  the  values 


If,  in  these,  A  equals  zero,  the  formula  reduces  to  (79),  which 
should  be  the  case,  as  the  whole  section  above  the  base  then 
becomes  a  single  trapezoid. 


ART.  35-]      DESIGN  OF  A   HIGH  ECONOMIC  SECTION.  1 1/ 

After  having  found  the  base  of  a  trapezoid  by  (88)  the 
value  of  s  should  be  computed  by  (85),  taking  tan  ^  =  o.  This 
will  be  at  first  greater  than  J#,  but  in  descending  lower  (usually 
before  d  becomes  100  feet)  a  trapezoid  will  be  found  where  ^ 
exceeds  \b.  As  soon  as  this  occurs  formula  (88)  ceases  to  be 
applicable,  for  the  section  has  not  a  sufficient  degree  of  sta- 
bility when  the  reservoir  is  empty.  The  back  must  now  be 
battered  so  that  s  shall  equal  ^b,  at  the  same  time  keeping 
t  =  ^b.  Introducing  these  two  conditions  into  (86)  and  solv- 
ing for  b  there  results 


which  gives  the  base  of  the  trapezoid,  and  thus  A^  becomes 
known.  The  amount  of  batter  required  is  now  found  by  in- 
serting in  (85)  the  value  of  s9  from  (81),  and  solving  for  tan  ipr 
namely  : 


(9o) 


in  which  s  is  to  be  taken  as  ^b.  Thus  the  trapezoid  is  fully 
determined,  and  the  next  one  can  be  designed,  taking  A^-\-  At 
as  the  new  Al  ,  s  as  the  new  s1  ,  and  b  as  the  new  a. 

After  having  found  the  base  of  a  trapezoid  by  (89),  the  com- 
pressive  unit-stress  at  the  ends  of  said  base  should  be  com- 
puted by  (80).  The  value  of  this  will  be  at  first  less  than 


1 1 8  MA  SONR  Y  DAMS.  [CHAP.  V. 

the  allowable  limit,  but  in  descending  lower  (usually  before  d 
becomes  150  feet)  a  trapezoid  is  reached  where  it  is  greater. 
As  soon  as  this  occurs  formula  (89)  ceases  to  be  applicable,  for 
the  base  of  the  section  has  not  sufficient  security  against 
crushing. 

The  next  value  of  b  is  to  be  derived  by  taking  t  as  given 
by  (82)  and  making  s  =  %b.  These  introduced  into  (86)  pro- 
duce a  quadratic  in  b,  and  this  will  be  used  until  the  com- 
pressive  stress  at  B  reaches  the  allowable  limit.  When  this 
occurs  s  must  be  made  greater  than  \b  by  expressing  its  value 
from  (58)  in  a  manner  analogous  to  (82).  The  two  values  of  s 
and  t  are  thus  stated  in  terms  of  S^  and  S,  the  limiting  unit- 
stresses  at  B  and  (7,  and  inserting  them  in  (86)  and  solving  for 
b  a  quadratic  is  found  from  which  all  the  remaining  trapezoids 
are  computed.  As  soon  as  any  b  is  found  Az  is  known,  and 
then  s  is  derived  by  (85),  taking  Al  -f-  A^  as  A.  Lastly,  using 
this  value  of  s,  the  batter  tan  ip  is  derived  by  (90). 

This  method  is  open  to  the  objection  that  the  formulas  of 
Article  20  do  not  probably  give  the  correct  law  of  distribution 
of  stress  on  the  base  of  polygonal  sections,  and  also  to  the 
objection  that  the  water  pressure  is  always  taken  as  horizontal 
in  direction.  On  the  other  hand,  it  has  the  advantage  of  being 
simple  in  use,  whereas  other  methods  but  little,  if  any,  more 
accurate  in  principle  lead  to  equations  of  high  degree  whose 
solution  can  only  be  effected  by  tentative  processes. 

By  the  help  of  this  method  the  engineers  of  the  Aqueduct 
Commission  of  the  city  of  New  York  deduced  an  economic 


ART.  35.]      DESIGN  OF  A   HIGH  ECONOMIC  SECTION. 


section  for  the  proposed  Quaker  Bridge  Dam.  The  top  thick- 
ness was  taken  at  20  feet  and  the  specific  gravity  of  the 
masonry  at  2.5.  The  following  are  results  for  the  theoretical 
section  to  a  depth  of  171  feet  (see  Table  II  in  Report  of  the 
Aqueduct  Commission,  1889). 


d 

b 

A 

tan  ^ 

t 

st 

s 

Si 

34-7 

20.0 

834 

O 

6.7 

10.  0 

13031 

6516 

50 

26.2 

1187 

O 

8.7 

10.5 

14156 

ii  328 

70 

37-4 

1823 

0 

12.5 

12.4 

15234 

15234 

90 

53-4 

2731 

0.115 

17.8 

17.8 

15984 

15984 

no 

71.2 

3977 

0.100 

25.2 

23.7 

16391 

17453 

130 

92.9 

5618 

0.170  ' 

35-1 

31-7 

16384 

18462 

150 

114.6 

7698 

0.170 

45-3 

40.1 

17078 

19930 

171 

137-4 

10339 

0.171 

56.1 

49.1 

18  219 

21  822 

In  this  table  the  first  column  contains  the  depth  of  the  water 
in  feet,  the  second  the  base  of  each  sub-trapezoid  in  feet,  the 
third  the  total  area  above  that  base  in  square  feet,  the  fourth 
the  batter  of  the  back,  the  fifth  and  sixth  the  distances  in  feet 
from  the  front  and  back  edges  of  the  base  to  the  lines  of 
resistance,  and  the  seventh  and  eighth  the  stresses  at  those 
edges  in  pounds  per  square  foot.  It  will  be  seen  that  the  San 
Mateo  dam,  170  feet  high  (Article  30),  has  about  61  per  cent 
more  material  than  this  economic  section  of  171  feet  height. 

Problem  35.  Design  an  economic  section,  taking  the  top 
thickness  as  30  feet  and  the  specific  gravity  of  the  masonry 
as  2\. 


I2O  MASONRY  DAMS.  [CHAP.  V. 


ARTICLE  36.    ADDITIONAL  DATA  AND  METHODS. 

There  has  now  been  given  snch  a  presentation  of  the 
theory  of  masonry  dams,  adapted  to  the  needs  of  students,  as 
will  serve  to  exemplify  the  principles  which  govern  their 
design.  A  few  concluding  remarks  concerning  data,  princi- 
ples, and  methods  will  now  be  made. 

The  force  of  the  wind  has  not  been  considered  in  the  data. 
If  the  wind  blows  up-stream  when  the  reservoir  is  filled,  the 
stability  of  the  dam  is  increased  ;  if  it  blows  down-stream,  its 
effect  will  be  to  produce  waves  rather  than  to  add  to  the 
water  pressure  on  the  back. 

The  pressure  due  to  the  impulse  of  waves  may  be  inferred 
from  the  fact  that  the  highest  pressure  observed  by  STEVEN- 
SON in  his  experiments  was  6100  pounds  per  square  foot.  The 
maximum  horizontal  pressure  per  linear  foot  on  the  top  of  a 
dam  from  wave  action  can,  therefore,  probably  not  exceed  this 
value  acting  over  three  or  four  feet  of  vertical  depth,  and  this 
only  when  the  reservoir  is  of  wide  extent. 

The  horizontal  pressure  at  the  water  line  due  to  the  thrust 
of  ice  should  be  taken,  in  the  opinion  of  a  board  of  experts 
on  the  Quaker  Bridge  Dam,  to  be  43  ooo  pounds  per  linear 
foot.  (Report  of  the  Aqueduct  Commission,  1889.) 

Let  H  be  the  horizontal  force  at  the  water  line  due  to  ice 
thrust,  or  wave  action.  Its  moment  will  be  Hd,  and  this  is  to 


ART.  36.]  ADDITIONAL  DATA  AND  METHODS.  121 
be  added  to  the  moment  of  the  water  pressure.  In  all  the 
preceding  formulas,  therefore,  the  quantity  —  should  be  re- 
placed by 1 —  fn  order  to  include  the  effect  of  this  hori- 

hg        vh 

zontal  force  in  the  computations.  For  instance,  if  the  example 
in  Article  32  is  to  include  the  effect  of  the  ice  thrust,  formula 
(84)  must  be  modified  as  stated,  taking  /^=  43000  pounds. 
Then  b  will  be  found  to  be  156.3  feet  instead  of  146.8,  and  the 
.area  of  the  trapezoid  will  be  about  5f  per  cent  greater  than 
before. 

When  the  computations  extend  below  a  permanent  water 
level  on  the  front  of  the  dam  the  effect  of  the  back  pressure 
can  easily  be  introduced  into  the  formulas  by  substituting 
d*  —  d*,  for  d\  where  d  is  the  depth  of  the  water  on  the  back 
of  the  dam,  and  */,  that  on  the  front. 

When  the  back  and  front  of  the  dam  are  covered  with 
earth  or  gravel  below  a  certain  level  its  action  may  be  approxi- 
mately estimated  by  computing  the  earth  pressures  according 
to  the  method  of  Article  8,  and  then  adding  the  moments  of 
these  to  the  other  external  moments.  Such  computations 
however,  will  always  be  liable  to  more  or  less  uncertainty,  and 
hence  should  be  made  with  caution. 

It  is  not  probable  that  the  theory  of  Article  20  gives  the 
correct  distribution  of  stress  on  the  wide  base  of  a  polygonal 
section,  and  it  seems  more  likely  that  in  such  cases  the  unit- 
pressures  at  the  ends  of  the  base  are  less  than  those  near  the 


122  MASONRY  DAMS.  [CHAP.  V.  ART.  36.} 

middle.  If  this  is  the  case,  the  formulas  probably  err  on  the 
side  of  safety,  even  though  they  neglect  the  influence  of  the 
shearing  stress  due  to  the  horizontal  pressures.  It  is  known 
(see  Mechanics  of  Materials,  Article  75)  that  a  shear  combines 
with  a  compression  normal  to  it  and  produces  in  another 
direction  a  greater  compression.  But  the  application  of  this 
principle  to  stresses  in  masonry  can  scarcely  be  made  until 
experimental  evidence  is  afforded  concerning  the  laws  of  dis- 
tribution of  the  unit-stresses. 

The  theory  of  a  dam  which  is  curved  in  plan  and  which 
acts  more  or  less  like  an  arch  has  not  been  considered  here. 
It  may  be  stated  as  the  general  consensus  of  opinion,  that  a 
section  which  resists  water  pressure  by  gravity  alone,  like 
those  designed  in  these  pages,  will  not  usually  be  rendered 
stronger  by  being  curved  in  plan.  A  curve,  however,  is  pleas- 
ing to  the  eye  and  impresses  the  observer  with  an  idea  of 
strength,  so  that  it  is  often  advisable  to  employ  it,  even  if  the 
length  of  the  dam  be  slightly  increased. 


AN  INITIAL  FINE  OF  25  CENTS 


~TCB~TS-19?4 

~JMTWmT 


W_H^494 


,'UN   2  5  1953 


JUN1 


REC'D  CD 


JAW  23  19SO 


LD  21-10m-5,'43 (6061s) 


